Light has momentum given by $E = pc$, which is derived from $E^2 =(m_0c^2)^2+(pc)^2$, where $m_0$ is set to zero. However I thought in this equation $p$ is defined as $p=\gamma m_0v$, which means it would also be zero if there was no rest mass. It seems like $E=pc$ is a postulate that happens to be true but isn't backed up by theory. Is there a more rigorous derivation of light's momentum?
Edited out $\gamma$
Relativistic momentum $p$ is the “spatial component of the particle 4-momentum”, whose magnitude reduces to $E/c$ for lightlike 4-momenta and $\gamma m v$ for timelike momenta.
I don't know how detailed for an explanation you are looking for, but maybe the following will be helpful.
The equation $p=\gamma m_0v$ is only valid for particles having mass. A photon thus has a momentum $p$ proportional to its wavelength $\lambda$
$$p=\frac{h}{\lambda}$$
where $h$ is Planck's constant. You might also know that a photon has an energy $E$ proportional to its frequency $f$ and thus proportional to its wavelength:
$$E=hf=\frac{hc}{\lambda}$$
(since $f\lambda=c$). It is thus easy to see that $$E=pc$$
If you are looking for a derivation why the above equations are true I'm afraid I can't help you – maybe someone else can answer in more detail.
