If $n^2 + 64n + 646 = m^2$ with $n$ and $m$ natural, then $n = \dfrac{-64 \pm \sqrt{64^2 - 4 \times (646 - m^2)}}{2}$. So $64^2 - 4 \times (646 - m^2) = 4 \times (378 + m^2)$ is a square number. This implies $378 + m^2$ is itself a square, say $378 + m^2 = k^2$. But $378 \equiv 2 \bmod 4$, while the difference of two squares must be $0$, $1$ or $3$ $\bmod 4$. This is a contradiction. Hence there are no solutions.