Sure, there are many ways to do this. Momentum is defined to be the derivative of the Lagrangian with respect to velocity, i.e. it describes how kinetic energy relates to relative velocity, as a function of rest mass - this is simply a generalisation of the good old p=mv, as we all know it from our school days:
\[p^{\mu } =\frac{\partial L}{\partial v^{\mu }} =mu^{\mu }\]
with the Lagrangian for relativistic motion being
\[L=-mc\int ds=-mc\int \sqrt{\eta _{\mu \nu } dx^{\mu } dx^{\nu }}\]
Energy, on the other hand, is defined as
\[E=p^{\mu } v_{\mu } -L\]
which is in essence a restatement of the fact that energy is the conserved quantity that arises from time-translation invariance in spacetime (see Noether‘s theorem for formal proof). Put these together and rearrange to get
\[E=\sqrt{m^{2} c^{4} +p^{2} c^{2}}\]
as stated above. You could also simply look at the general form of a 4-momentum vector, and see immediately that its temporal and spatial parts are related as above. This would be a standard way to derive this, but there are many, many other ways to do this, both informally and in very formal ways.
For massive particles at rest you have m<>0 and p=0 - insert into the above to get
\[E=mc^2\]
as requested. You can also derive all of this from first principles - the symmetries of Minkowski spacetime - in a very formal way by using Noether‘s theorem; this gives you the energy-momentum tensor as a conserved quantity, and from its vanishing divergence you can derive the above expression as well.
The problem here isn‘t any of the above, because you can find all of this in pretty much any standard undergrad text on Special Relativity. This is basic stuff, and it is not in contention. The problem though is that your response to this will be along the lines of: “You don’t understand anything, you are just parroting what you have read!”.
Am I right?