From the given condition, $9(m^2+1) = (n+m)(n−m)$. Note that $m^2 + 1$ is a prime number not equal to 3. Hence there are four cases. a) $n − m = 1, n + m = 9(m^2 + 1)$. Subtraction yields $9m^2 + 8 = 2m$, which is impossible. b) $n − m = 3, n + m = 3(m^2 + 1)$. Subtraction yields $3m^2 = 2m$, which is impossible. c) $n − m = 9, n + m = m^2 + 1$. Subtraction yields $2m = m^2 −8$, so that $m = 4$ and $n = 13$. Note that $m^2 + 1 = 17$ is indeed a prime number. d) $n − m = m^2 + 1, n + m = 9$. Subtraction yields $−2m = m^2−8$, so that $m = 2$ and $n = 7$. Note that $m^2 + 1 = 5$ is indeed a prime number. In summary, there are two solutions, $(m, n) = (2, 7)$ or $(4, 13)$.