$n = 256$. Such $n$ as described in the problem satisfies $n = 2p^3 + p + 1 = q^k$, where $k$ is some positive integer. Obviously $2p^3 + p + 1$ is even, so $q = 2$.

It is easy to check that $3|2p^3+p$ must hold. Thus from $3|2^k-1$ we know $2|k$. Let $A = 2^{k/2}$, we have $(A+1)(A-1) = 2p^3+p = (2p^2+1)p$, so $p|(A-1)$ or $p|(A+1)$. Let $A \pm 1 = pt$, where $t$ is a positive integer. Then we have
$$pt^2 \pm 2t = t(pt \pm 2) = 2p^2 + 1.$$
We have $p|\pm2t+1$, so $2t+1 \ge p$. Thus $8t^2+8t+3 \ge 2p^2+1 \ge pt^2-2t$.
From this we get $t < 5$, or when $t \ge 5$, $11t^2 > 8t^2+10t+3 \ge pt^2$, that is $p < 11$. If $t < 5$, then from $2t+1 \ge p$ we also have $p < 11$.
Thus the only remaining possibilities are $p = 3, 5, 7$. Substituting and checking, we know $p = 5, k = 8, n = 256$ is the unique solution.