No lol, there's a momentum component to that equation which is just conveniently 0 for massive objects at rest, photons don't have a rest reference frame and are governed by E=pc.

nope, the full relativistic energy relation is:

E^2^ = m^2^ c^4^ + p^2^ c^2^

p is the momentum and for a (massive) particle at rest the momentum p=0, taking the square root if both sides you get the more familiar:

E = m c^2^

Now a photon (and any other massless particle) can't be at rest, it is forced to always travel at the speed of light and since it is massless m=0 and the energy becomes (again taking the square root):

E = p c

When a particle has both mass and isn't at rest (but not traveling anywhere close to the speed of light) the E = m c^2^ is much, much larger than the E = p c part (ignoring that a square root isn't linear {Hello Dirac}). Because the speed of light c is such a a huge number that squaring it makes it even bigger. It is usually fair to say that mass is 'equivalent' to energy, but it isn't strictly true and actually false for massless particles (or particles traveling close to the speed of light [velocity v~c] -> p=m v~m c -> E = pc ~ m c^2^ , which has close to c^2^ in it).

So photons have energy, not because they have mass (where massive particles have most of their energy), but because they have momentum (p).

You bring up the theoretical black hole from photons, which are called 'Kugelblitz' black holes, iirc. They (theoretically could) exist, not because photons have some sort if mass, but because spacetime curves because of the energy content, not mass. Again, for most regular objects, the vast majority of its energy comes from its mass and the momentum doesn't play a huge role. But for photons all their energy comes from their momentum, since they don't have any mass.

Source: my bachelors degree in physics, I suppose.