We have $p^2 = n^2 - m^2 = (n - m)(n + m)$. Since $p$ is a prime, the number $p^2$ has the divisors $1$, $p$ and $p^2$. Since the two factors $n - m$ and $n + m$ are distinct, they cannot be both equal to $p$. Furthermore, $n - m$ is smaller than $n + m$, therefore, $n - m = 1$, i.e. $n = m + 1$.

We find
$$p^2 + m^2 = (m + 1)^2 \iff p^2 = 2m + 1.$$

This immediately implies that $p$ is odd, therefore $p \ge 3$. We find $2m+1 = p^2 \ge 3p > 2p+1$,
which gives $m > p$ as desired.