The required integers are $k = 3$ and $n = 2$. It is readily checked that these integers satisfy the condition in the statement.

To show that there are no other such, write $2^k + 10n^2 + n^4 = N^2$, where $N$ is a positive integer, so $(N - n^2 - 5)(N + n^2 + 5) = 2^k - 25$.

The latter shows that $N \ge n^2 + 6$ if $k \ge 5$, and of the first four positive integers, only $k = 3$ yields a positive integer $n = 2$, in which case $N = 8$.

Henceforth, let $k \ge 5$, and notice that $n$ must be even, for otherwise $N^2 = 2^k + 10n^2 + n^4 \equiv 3 \pmod 4$, which is impossible. Let $2^m, m \ge 1$, be the highest power of $2$ dividing $n$.

If $k > 2m+1$, then $2^{2m+1}$ is the highest power of $2$ dividing $2^k+10n^2+n^4 = N^2$, which is again impossible.

Consequently, $k \le 2m+1$, so $2^k \le 2^{2m+1} \le 2n^2$, and $N^2 = 2^k + 10n^2 + n^4 \le 12n^2 + n^4 < (n^2 + 6)^2$; that is, $N < n^2 + 6$, in contradiction with $N \ge n^2 + 6$ established above. This ends the proof.