The reason that there are only two polarization states not three as you'd expect from a vector field is that the photon is massless.
It goes back to the representation theory of the Poincare group. Looking for irreducible unitary representations you come to the conclusion, that good states are the momentum eigenmodes. In order to be irreducible, the Lorentz transformations must bring you from any "standard momentum" to any other. This implies that ##p \cdot p=m^2## with ##m^2## a given constant for all states represented by the irrep. Physically relevant are the cases ##m^2>0## (leading to the many-body description of massive particles, when the field is quantized) and ##m^2=0## (massless particles).
For massless particles you can choose the standard momentum as ##p=\Lambda(1,0,0,1)## (with ##\Lambda \neq 0## arbitrary). The irrep. then is further characterized completely by the representation of the subgroup which leaves this standard momentum invariant (the socalled "little group"). The group leaving the light-like standard momentum invariant is not compact since this group is equivalent to ISO(2), i.e., the group built by translations and rotations of the Euclidean plane. Now, to have only discrete intrinsic ("polarization") degrees of freedom, you must represent the translations in this plane trivially and only the rotations non-trivially (if you represent them also trivially, you get massless spin-0 particles, which is mathematically fine but not observed in nature). This leads to the conclusion that the only remaining degree of freedom is the rotation around the ##z##-axis, also leaving the light-like standard vector unchanged. This is a U(1), and it's representation is given by multiplication with a phase factor ##\exp(-\mathrm{i} \lambda \varphi)##. In principle, ##\lambda \in \mathbb{R}##, can take any value, but you want to lift the representation of the little group to one of the entire Poincare group, and there you have the usual rotations in space, and thus you must have ##\lambda \in \{0,\pm 1/2,\pm 1,\ldots \}##. So for massless particles you have only two polarization-degrees of freedom, which are represented by the rotations around the direction of the momentum of the massless particle, whose generator is the projection of the particle's angular momentum to the direction of its spin, the socalled helicity. For a massless spin-1 particle you the two helicities, ##\lambda=\pm 1##. In classical electromagnetism this describes left- and right-circular polarized electromagnetic waves.
