All integers $n = 3^k$ with $k \in \mathbb{Z}^+$ satisfy $n \mid 2^n + 1$. Indeed, since $3 \mid 2 + 1$, by the lifting the exponent lemma, we have
$$v_3(2^{3^k} + 1^{3^k}) = v_3(2 + 1) + v_3(3^k) = 1 + k.$$
This implies $3^{k+1} \mid 2^n + 1$, and hence $n \mid 2^n + 1$.

The only prime number $n$ satisfying $n \mid 2^n + 1$ is $n = 3$. Indeed, let $n = p$ be a prime. By the Fermat little theorem, we have
$$2^p + 1 \equiv 2 + 1 = 3 \pmod{p}.$$
This is congruent to $0$ modulo $p$ if and only if $p \mid 3$, i.e. $p = 3$.