The problem is equivalent to finding the number of solutions in positive integers to the following system of equations: \[ \begin{cases} a^2 + b^2 = c^2 \\ ab = (a + b + c)p^m\end{cases} \] Denoting $ a + b = t$, we have $ t^2 = c^2 + 2(t + c)p^m$ implying that $ (t + c)(t - c) = 2(t + c)p^m$ and \[ c = t - 2p^m. \] Therefore, the original system is equivalent to \[ \begin{cases} a + b = t \\ ab = 2(t - p^m)p^m\end{cases} \] implying that $ a, b$ are the roots of the following quadratic equation: \[ x^2 - t x + 2(t - p^m)p^m = 0 \] which is solvable in integers only if its discriminant $ t^2 - 8 (t - p^m)p^m = (t - 4p^m)^2 - 8p^{2m}$ is a square. If $ (t - 4p^m)^2 - 8p^{2m} = q^2$ then $ 8p^{2m} = (t - 4p^m - q)((t - 4p^m + q)$. All possible non-negative such $ q$ are obtained as $ q = \frac {d_2 - d_1}{2}$ where $ d_1 d_2 = 8p^{2m}$ and $ d_2 \geq d_1$ are of the same oddness. If $ p = 2$ then $ 8p^{2m} = 2^{2m + 3}$ and $ q = \frac {2^k - 2^{2m + 3 - k}}{2} = 2^{k - 1} - 2^{2m + 2 - k}$ where $ k = m + 2\dots,2m + 2$. These $ m + 1$ distinct values of $ k$ give the following solutions: \[ \begin{cases} a = 2^{k - 1} + 2^{m + 1} \\ b = 2^{2m + 2 - k} + 2^{m + 1} \\ c = 2^{k - 1} + 2^{2m + 2 - k} + 2^{m + 1}\end{cases} \] If $ p > 2$ then $ q = \frac {4p^k - 2 p^{2m - k}}{2} = 2 p^k - p^{2m - k}$ (for $ k = m,\dots,2m$) or $ q = \frac {2p^k - 4 p^{2m - k}}{2} = p^k - 2 p^{2m - k}$ (for $ k = m + 1,\dots,2m$). Therefore, the number of solutions here is $ 2m + 1$. Explicit solutions can be derived as above.

There is a MUCH simpler proof knowing that all Pythagorean triples can be parametrized as (k(m^2-n^2), 2mnk, k(m^2+n^2) where m, n, and k are positive integers such that m - n is odd and gcd(m, n) = 1. I won't write it here because I don't know how to use Latex and it will look horrible, but that idea just about solves the problem instantly.