Suppose the contrary, and let $m+n-1=p$ for some $p\ge 3$. Then $m+n\equiv 1\pmod{p}$, forcing $m^2+n^2-1\equiv -2mn\pmod{p}$. If $p\mid m^2+n^2-1$, then $p\mid 2mn$. As $p>2$, this is possible only when $p\mid m$ or $p\mid n$. Notice, however, that as $p=m+n-1>\max\{m,n\}$ (since $m,n\ge 2$), this cannot be true.