The issue is not as simple as suggested by the answers to the OP. Whether four numbers (I use units with ##c=1## for convenience) ##(\omega,\vec{k})## transform like the components of a four-vector is not so clear from just writing these four components down. It's a question of context. The notation suggests that it is frequency and wave vector of some wave. Thus there is some wave equation describing these waves. This implies that there's a dispersion relation dependent on the wave equation. If the wave equation can be written in Lorentz covariant form this dispersion relation must be covariant, and the only invariant you can built of four numbers ##(\omega,\vec{k})## assumed (sic!) to be components of a four-vector without further physically relevant vectors or tensors around (i.e., a wave propagating in vacuum) is ##k_{\mu} k^{\mu}=m^2## with a scalar quantity ##m^2 \geq 0##. Since this is a Lorentz-invariant expression supposed the four numbers ##(\omega,\vec{k})## transform like four-vector components, you have a Lorentz-covariant dispersion relation.
This implies that the phase factor, which is part of the mode decomposition of the field with respect to momentum eigenmodes (where momentum is defined as the Noether charge under spatial translations), ##\exp(-\mathrm{i} x_{\mu} k^{\mu})## is a scalar either, as is immediately clear by the notation since by assumption the ##k^{\mu}## and the ##x^{\mu}## transform as comonents of four-vectors under Lorentz transformations.
Whether or not a theory makes sense within relativistic physics is, whether you can find a transformation law for all involved quantities under the symmetry transformations underlying the (special) relativistic space-time structure, i.e., it must be possible to define transformation rules of the fields under consideration (and the natural way to express relativistic natural laws is in terms of local (quantum) field theory) realizing the proper orthochronous Lorentz group.
The most simple idea are scalar, vector and arbitrary tensor fields. As QT tells us one can also consider spinor representations of the Poincare group.
The most simple example is, of course, just a scalar field, and a nice covariant equation is the Klein-Gordon equation,
$$(\Box + m^2) \phi(x)=0.$$
With the ansatz ##\phi(x)=\phi_0 \exp(-\mathrm{i} k_{\mu} x^{\mu})## you get the above postulated dispersion relation, and everything is manifestly covariant.
For electromagnetism it's somewhat more complicated, but you can realize it in terms of the four-vector potential with components ##A^{\mu}##, assumed to transform under Lorentz transformations as four-vector fields. The physical quantities are of course the electric and magnetic field components, of put in the usual real-valued relativistic formulation the Faraday tensor components
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.$$
Then the equation of motion reads
$$\partial^{\mu} F_{\mu \nu}=j_{\nu},$$
where ##j^{\nu}## are the components of the electric-charge four-current ##(\rho,\vec{j})##.
Since this doesn't uniquely determine ##A_{\mu}##, because the physical quantities ##F_{\mu \nu}## don't get changed when adding an arbitrary four-gradient field (gauge invariance), i.e., the two fields ##A_{\mu}## and ##A_{\mu}'=A_{\mu} + \partial_{\mu} \chi## are equivalent for any scalar field ##\chi##, you can impose an arbitrary gauge-fixing condition. A particularly convenient and manifestly covariant one is the Lorentz-gauge condition
$$\partial_{\mu} A^{\mu}=0.$$
With that condition the equation of motion becomes
$$\Box A^{\mu}=j^{\mu}.$$
For free fields, i.e., ##j^{\mu}=0##, you can impose even one more constraint, like, e.g., ##A^0=0##, and you see that the plane-wave solutions are transverse vector fields with the dispersion relation given by ##k_{\mu} k^{\mu}=0##.
Since everything is manifestly covariant, assuming that ##A^{\mu}## and ##j^{\mu}## transform as components of four-vector fields, you have a proper relativistic field theory.
