Infinitely many n dividing 2ⁿ + 1

Base case. $k=0$: $3^1\mid2^1+1=3$ ✓.

Induction step. Assume $3^{k+1}\mid2^{3^k}+1$, and let $x=2^{3^k}$. Then
$$2^{3^{k+1}}+1=x^3+1=(x+1)(x^2-x+1).$$
The first factor is divisible by $3^{k+1}$. For the second, $x\equiv-1\pmod3$, so $x^2-x+1\equiv1+1+1=3\equiv0\pmod3$. Therefore $3^{k+2}$ divides the product.

Conclusion. So $3^{k+1}\mid2^{3^k}+1$, in particular $3^k\mid2^{3^k}+1$: with $n=3^k$, $n\mid2^n+1$. There are infinitely many such $n$.

Check. $k=2$: $2^9+1=513=27\cdot19$ ✓.

Lesson. To gain one more power of a prime, factor $x^p\pm1$ and show that the cofactor is divisible by $p$ exactly once.