Without losing generality we may assume $m \geqslant n$ Suppose that $m+n = p+1$ where $p$ is a prime. Since $m^2 + n^2 - 1 \equiv 0 \pmod p$ and $m^2 \equiv (n-1)^2 \pmod p$, we have $(n-1)^2 \equiv (n^2 - 1) \pmod p$ and so $p \mid 2n-2$, which means that either $p = 2$ or $p \mid n-1$, both of which are impossible. Hence $m+n-1$ cannot be prime.

Let's assume $m+n-1=p$ where $p$ is prime number $(m+n-1) | (m+n-1)(m+n+1) $ $(m+n-1) | 2mn$ we know that $p \ne 2$ with condition then $\gcd(m,m+n-1)=1$ $\gcd(m+n-1,n)=1$ contradicition.