Austrian Mathematical Olympiad

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 $$p^2+m^2=(m+1{)}^2⟺p^2=2m+1.$$

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.