Problem 1064

$n$ can be $47$ or $17$.

Suppose $n^2 + 32n + 8 = m^2$ where $m$ is a positive integer. Note that this can be rewritten as $(n+16)^2 - 248 = m^2$, which implies
$$(n + m + 16)(n - m + 16) = 248 = 2^3 \times 31.$$
Hence, $(n+m+16, n-m+16) = (248, 1), (124, 2), (62, 4), (31, 8)$. For each pair, we can solve the system of two linear equations in two unknowns. The only positive integer solutions are $(n, m) = (47, 61), (17, 29)$.