23. Let us assume $n > 1$. Obviously $n$ is odd. Let $p \geq 3$ be the smallest prime divisor of $n$. In this case $(p-1, n)=1$. Since $2^n + 1 \mid 2^{2n} - 1$, we have that $p \mid 2^{2n} - 1$. Thus it follows from Fermat's little theorem and elementary number theory that $p \mid (2^{2n} - 1, 2^{p-1} - 1) = 2^{(2n, p-1)} - 1$. Since $(2n, p-1) \leq 2$, it follows that $p \mid 3$ and hence $p = 3$. Let us assume now that $n$ is of the form $n = 3^k d$, where $2, 3 \nmid d$. We first prove that $k = 1$. Lemma. If $2^m - 1$ is divisible by $3^r$, then $m$ is divisible by $3^{r-1}$. Proof. This is the lemma from (SL97-14) with $p = 3, a = 2^2, k = m$, $\alpha = 1$, and $\beta = r$. Since $3^{2k}$ divides $n^2 \mid 2^{2n} - 1$, we can apply the lemma to $m = 2n$ and $r = 2k$ to conclude that $3^{2k-1} \mid n = 3^k d$. Hence $k = 1$. Finally, let us assume $d > 1$ and let $q$ be the smallest prime factor of $d$. Obviously $q$ is odd, $q \geq 5$, and $(n, q-1) \in \{1, 3\}$. We then have $q \mid 2^{2n} - 1$ and $q \mid 2^{q-1} - 1$. Consequently, $q \mid 2^{(2n, q-1)} - 1 = 2^{2(n, q-1)} - 1$, which divides $2^6 - 1 = 63 = 3^2 \cdot 7$, so we must have $q = 7$. However, in that case we obtain $7 \mid n \mid 2^n + 1$, which is a contradiction, since powers of two can only be congruent to 1, 2, and 4 modulo 7. It thus follows that $d = 1$ and $n = 3$. Hence $n > 1 \Rightarrow n = 3$. It is easily verified that $n = 1$ and $n = 3$ are indeed solutions. Hence these are the only solutions.