If the particle has zero rest mass, like a photon, it still has energy. E^2 = p^2 c^2 + m_0^2 c^4 captures that correctly, while E = mc^2 where 'm' is the relativistic mass, does not.
No, it does correctly capture that. The relativistic mass of a photon is not zero even though its rest mass is. However for a zero-mass particle you cannot use the Lorentz transformation to calculate relativistic mass, because v = c and so you would be dividing zero by zero.
You can correctly calculate the relativistic mass of a photon in other ways, such as from its momentum or its gravitational interaction, and doing so gives you the energy you'd expect from E = mc².
Oh heavens whatever caused you to believe this? The momentum p of a photon is given by p=h/λ where h is the Plank Constant and λ is the wavelength. Compton won a Nobel prize for this.
Of course p = h/λ. Thus the relativistic mass m is given by p/v = p/c = h/cλ. And a photon's energy is given by E = hc/λ. This is all in agreement with the relativistic mass you'd calculate from the photon energy E = mc².
Except that it does gravitate more. Avoiding the confusion that it doesn't is one of the conceptual benefits of relativistic mass. Gravitation is dependent on the stress-energy tensor, which contains... drum roll... relativistic mass, not rest mass!
For that matter, you've already noticed that the Lorentz transform becomes nonsensical as v → c when starting from rest mass, whereas the transform works just fine when using relativistic mass as the starting point, and nothing blows up, and you can avoid any confusion.
