Solution:

We may limit the search to nonnegative integers $n$, since $n^{2}+15 = (-n)^{2}+15$. Suppose there is a nonnegative integer $m$ such that
$$\begin{aligned} n^{2}+15 & = m^{2} \\ 15 & = m^{2} - n^{2} \\ & = (m+n)(m-n) . \end{aligned}$$
Note that the factors $m+n$ and $m-n$ are not both negative, since their sum $2m$ is nonnegative, so they are both positive. Now the only ways to factor $15$ into two positive integer factors are $15 \cdot 1$ and $5 \cdot 3$. Also $m+n \geq m-n$ since $n \geq 0$.

If $m+n=15$ and $m-n=1$, then
$$n = \frac{(m+n)-(m-n)}{2} = \frac{15-1}{2} = 7$$
and $m=8$.

If $m+n=5$ and $m-n=3$, then
$$n = \frac{(m+n)-(m-n)}{2} = \frac{5-3}{2} = 1$$
and $m=4$.

Both of these solutions indeed work in the original problem. Finally, allowing for negative $n$, we compute the answers to be $-7, -1, 1$, and $7$.