It is stated that according to quantum field theory the value $0$ for the projection of the spin of the photon would be impossible because the photon has null mass, thus photon has two polarizations only (unlike $W$ and $Z$ which have 3 polarisations).
Could it be explained further what is the relationship/explanation with the argument of null mass for impossibility of projection 0 of spin ? (I do know that photon has null mass).
Is there an easy mathematical proof ?
The full proof is based on carefully constructing unitary representations of the Poincaire group. This is done, for example, in chapter 2 of Weinberg's quantum field theory textbook (vol 1).
Here's a short, not-rigorous, physical explanation. For any particle the states will be labeled by the momentum, plus the spin degrees of freedom. For a massive particle, we can go to the rest frame where the spin is zero, and then the spin degrees of freedom are described by the representation theory of the $SO(3)$ group (ie, angular momentum) which is done in undergraduate quantum mechanics. In particular, a particle with spin $s$ will have $2s+1$ degrees of freedom. For a massless particle, we will not be able to use the same argument and so we would expect the answer to be different. Roughly speaking, what happens is that the spin degrees of freedom of a massless particle effectively behave like the spin degrees of freedom of a massive particle in one smaller dimension (we lose the dimension along the direction of propagation). So in the Standard Model, the spin degrees of freedom of a massless particle are determined by the rotation group $SO(2)$, which is the little group for massive particles in $2+1$ dimensions. In fact, the little group for massless particles in $3+1$ dimensions turns out to be $ISO(2)$, or the symmetries of the Euclidean plane (rotations plus a translation). (Thanks to Libretarian Feudalist Bot in the comments for correcting my original answer).
To really explain this requires a much more detailed argument, but this answer is meant to give you a flavor, and point you to Weinberg who has the definitive (but not easy!) treatment.
