Suppose I have a photon with energy $E$ (in its rest frame) moving along the $z$ direction with momentum $\mathbf{p} = p_z\mathbf{\hat{z}}$:
The Lorentz transform in the laboratory frame should be: $$ E'=\gamma(E-p_zv) $$
so using $p_z = E / c$ and $v=c$ for light, I get that $E'=0$.
How is this possible? Is there a physical meaning?
What you are calculating is the invariant mass, for a photon the invariant mass is 0. This transformation is only 1 component of the Energy-Momentum 4-vector. Use the full form and you will get the result you want. The full form is given right at the end of this page
It is incorrect to state that there is a photon with energy $E$ in its rest frame, but it would be correct to talk about a photon with energy $E$ in some frame. In this case, the equation that you wrote would be valid to express the energy of that photon observed from a different frame (that you choose to characterize with a prime) that moves away with normalized velocity $\beta=\frac{v}{c}$,
$\displaystyle E^\prime = E \, \gamma \, ( 1 - \beta ) = E \, \sqrt{\frac{1 - \beta}{1 + \beta}}$.
This expression follows from exactly what you wrote. This result is the well known expression of the Doppler effect, and is valid for any $\beta < 1$. In the case of $\beta = 1$, you have found that $E^\prime = 0$. You cannot have an observer moving at the speed of light with respect to any other observer, so no one will ever measure $E^\prime = 0$ anyway. The somehow paradoxical idea that some observer would measure a zero energy is protected by the fact that no observer can ever reach the necessary speed for this to happen. However, you can think of a reference frame in the mathematical sense, moving at that speed. In this case your conclusion would be the same that it would be for a massive particle: in the rest frame where that particle is at rest, the energy is the rest energy, i.e. its mass (times $c^2$). In the (non-physical) frame with the photon being at (non-physical) rest, the energy of that photon would be its mass, i.e. $0$.
