When can 2ⁿ + 1 or 2ⁿ − 1 be prime?

1. Suppose $n$ is not a power of $2$: $n=2^s m$ with $m\ge3$ odd. Let $x=2^{2^s}$. Then $2^n+1=x^m+1$ is divisible by $x+1$, and $1<x+1<x^m+1$. So $2^n+1$ is composite. Hence $n$ must be a power of $2$ (the numbers $2^{2^k}+1$ are the Fermat numbers).

2. Suppose $n=ab$ with $a,b\ge2$. Then $2^n-1=(2^a)^b-1$ is divisible by $2^a-1$, and $1<2^a-1<2^n-1$. So $2^n-1$ is composite. Hence $n$ is prime (the numbers $2^p-1$ are Mersenne numbers).

Remark. The converses are false: $2^{11}-1=23\cdot89$ and $2^{32}+1=641\cdot6700417$.

Lesson. Whenever an exponent has a divisor, a geometric-sum factorisation $x^m\pm1$ produces a proper factor.