Masseless particles are different from massive ones in their behavior under Poincare transformations. One can understand this only by a careful study of the unitary representations of the Poincare group, see e.g. my manuscript on QFT about it:
http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
Here, I try to give a short qualitative explanation. Let's start with massive particles. The analysis of the Poincare-group representation lead to a one-particle basis in Fock space given by mass-momentum eigenstates, and spin. The spin is determined by the behavior of the zero-momentum states under rotation. For a massive spin-1 particle these rotations (the "little group") are represented by the s=1 representation, and that means the possible values for the 3-component of the spin are s_z=\{-1,0,1\}. That's pretty easy to understand since we are familiar with this already from non-relativistic quantum theory of particles with spin 1 (although usually one treats only particles with spin 0 and 1/2, the former because it's the most simple case, the latter because a broad field of applications for non-relativistic quantum mechanics is atomic, molecular, and condensed-matter physics, where electrons play a very important role, and those are spin-1/2 particles).
The case of massless particles is a bit more complicated. This is due to the fact that first of all there is no restframe for a massless particle. In any inertial frame, it moves with the speed of light, which means that its four-momentum is light-like (i.e., a null vector of the Minkowski metric). Thus the three momentum cannot be 0.
Now you can always take a frame of reference, where the momentum points in a certain direction (usually one takes the z axis for this "standard momentum"). To further classify the representation, you have to figure out which Lorentz transformations keep the standard momentum fixed. Of course, that's the rotations around the direction of the standard momentum, but there are also two more independent one-parameter subgroups. All these transformations together give a group that is isomorphic to ISO(2), i.e., the group of translations and rotation in the two-dimensional Euclidean plane. This group is, contrary to the rotation group (or better said its covering, the SU(2)) not compact, and as is well known from quantum mechanics, the translations give continuous degrees of freedom, namely the "eigenvalues" of the momentum components which are the real numbers for each component. In our case, this would mean that we had an infinite number of spin-like states. A particle with such a strange property hasn't been observed ever. Thus, to discribe massless particles realistically, we have to make sure that the "translations" (in our context better named "null-rotations", because one rotates around a space-like four-momentum) are represented by the trivial representation. If one thinks in terms of classical fields, this leads precisely to the U(1) gauge property of electrodynamics, i.e., massless spin-1 particles are necessarily represented as gauge bosons! The remaining "spin-like degrees of freedom" thus come only from the rotations around the direction of the standard momentum, which is an SO(2) ~ U(1) group. Then, to get the rotations for all three momenta right, one comes to the conclusion, that one can only use the rotations around the standard-momentum direction as factors exp(i \lambda phi) with \lambda \in \{0,\pm 1/2,\pm 1,\ldots \}. Of course \lambda is the helicity of the particle. For massive particles this quantity refers to the projection of the spin to the direction of its momentum, but as is clear from the above considerations, for massless particle the concept is somewhat more subtle (see my manuscript for a quite detailed explanation).
Thus a massless particle with spin s \geq 1/2 has two spin-like degrees of freedom, no matter of the value of s. For s=1 that means it has only two helicity states, namely \lambda=\pm 1. A massive spin-1 particle has three spin-degrees of freedom according to the possible values of s_z \in \{0,\pm 1\}.
Now another issue is the case of massive gauge bosons that became their mass through the Higgs mechanism. In the case of non-Abelian gauge symmetries that's the only way to build Dyson-renormalizable models with massive spin-1 fields, and the electroweak part of the standard model rests upon this discovery by Higgs, Kibble, Anderson, Brout,...
The Higgs mechanism works as follows: You introduce a multiplet of the gauge group in question and give it a negative mass and a quartic self-interaction such that the stable vacuum state is given at a finite vacuum expectation value of this field. If the symmetry under consideration is a global symmetry, you get some massless boson-degrees of freedom, the famous Nambu-Goldstone bosons and some massive particles left. This happens in the chiral limit of the light-quark part in QCD. How many Goldstone bosons and how many massive particles you get in the model depends on the dimension of the group and the dimension of the subgroup that leaves the stable-vacuum field configuration invariant and the dimension of the representation (as a real Lie group!). Take only u and d quarks, which (in the chiral limit) are assumed to be massless. Then one has a symmetry group \mathrm{SU}(2)_L \times \mathrm{SU}(2)_R. To describe pions, one can introduce four real scalar fields to represent this ciral group as SO(4) and write down an SO(4)-symmetric lagrangian with negative mass-squared and a four-particle point interaction. That's the linear \sigma model of chiral symmetry. The non-zero vacuum expectation value defines then a direction in the four-dimensional field space, and the remaining symmtry group are the rotations around this axis. Thus the subgroup that leaves the stable vacuum invariant is SO(3). That means, we have 3 massless Goldstone bosons (the pions in the chiral limit). Since we started with a four-dimensional representation, there's one massive boson left, which represents the \sigma meson, which however is a broad resonance rather than a well-defined particle.
If the symmetry is a local gauge symmetry, you can find a special special gauge, the unitary gauge, where all those fields which would make massless Goldstone bosons in the case of a local symmetry, are absorbed into some of the gauge bosons of the group. Each of these gauge bosons become massless. The other bosonic degrees of freedom become also massive and these are the Higgs bosons of the model. In the electroweak standard model the gauge group is G=\mathrm{SU}(2) \times \mathrm{U}_1. The group is four-dimensional. In the version with a "minimal Higgs sector" one introduces a doubled of complex scalar fields (i.e. 4 real field-degrees of freedom). The symmetry is broken to the electromagnetic H=U(1) (which is different from the U(1) we started with!). Counting according to our rules leads to dim(representation of the gauge group to which the scalar fields belong)-dim(H)=4-1=3 would-be Goldstone bosons, which are absorbed to three of the four gauge-boson degrees of freedom, i.e., we end up with 3 massive vector bosons and one massless vector boson, which become the W/Z bosons and the photon, respectively. The photon stays massless as it must be. It is of course not a Goldstone boson as claimed in one earlier posting!
