Throughout the solution $n$ stands for a positive integer. By Euler's theorem, $\left(2^{3^{n}}+1\right)\left(2^{3^{n}}-1\right)=2^{2 \cdot 3^{n}}-1 \equiv 0\left(\bmod 3^{n+1}\right)$. Since $2^{3^{n}}-1 \equiv 1(\bmod 3)$, it follows that $2^{3^{n}}+1$ is divisible by $3^{n+1}$.

The number $\left(2^{3^{n+1}}+1\right) /\left(2^{3^{n}}+1\right)=2^{2 \cdot 3^{n}}-2^{3^{n}}+1$ is greater than 3 and congruent to 3 modulo 9, so it has a prime factor $p_{n}>3$ that does not divide $2^{3^{n}}+1$ (otherwise, $2^{3^{n}} \equiv-1$ (mod $\left.p_{n}\right)$, so $2^{2 \cdot 3^{n}}-2^{3^{n}}+1 \equiv 3\left(\bmod p_{n}\right)$, contradicting the fact that $p_{n}$ is a factor greater than 3 of $2^{2 \cdot 3^{n}}-2^{3^{n}}+1$).

We now show that $a_{n}=3^{n} p_{n}$ satisfies the conditions in the statement. Since $2^{a_{n}}+1 \equiv$ $2^{3^{n}}+1 \not \equiv 0\left(\bmod p_{n}\right)$, it follows that $a_{n}$ does not divide $2^{a_{n}}+1$.

On the other hand, $3^{n+1}$ divides $2^{3^{n}}+1$ which in turn divides $2^{a_{n}}+1$, so $2^{3^{n+1}}+1$ divides $2^{2^{a_{n}}+1}+1$. Finally, both $3^{n}$ and $p_{n}$ divide $2^{3^{n+1}}+1$, so $a_{n}$ divides $2^{2^{a_{n}}+1}+1$.

As $n$ runs through the positive integers, the $a_{n}$ are clearly pairwise distinct and the conclusion follows.