If $\exists n \in \mathbb{Z+}$,s.t. $2^n + 1$ is a prime, prove $\exists k \in \mathbb{Z+}, n = 2^k$ The actual description of the question is: 
Suppose that $2^n + 1$ is a prime where $n$ is a positive integer. Prove that
$n = 2^k$ for some positive integer $k$.
There is only one approach that know of:  To show that there is factorization in which at least one factor $(x-a)=1$, and the others are prime. A nice simplification would be to show that all except one root is having value such that the factor has value = $1$.   For example, if given quadratic: $x^2 + 2x - 3 = (x+3)(x-1)$. As have two factors, so one of these factors to be $1$, and the other prime. 
This leads to: $$x-1=1 \implies x=2 \implies (x+3)(x-1) = 5$$
else, $$x+3=1 \implies x = -2 \implies (x-1)(x+3) => -3$$
But, if consider positive integers then $x=2$ is the value.
if given quadratic: $x^2 - 5x +6 = (x-3)(x-2)$. As have two factors, so one of these factors to be $1$, and the other prime. 
This leads to: $$x-3=1 \implies x=4 \implies (x-3)(x-2) = 2$$
else, $$x-2=1 \implies x = 3 \implies (x-3)(x-2) => 0$$
But, if consider that $0$ is not a prime, then $x=4$ is the value.
So, let us try to work out cases with value of $k$ starting form $0$.
$k=0, 2^0=1, 2^1+1 =3$
$k=1, 2^1=2, 2^2 +1=5$
So, weak induction will suffice here, as strong induction is not needed as no need for previous values to prove new case.
So, let the assumption be that it holds for $k=l$, need prove for $k=l+1$,  but, $2^{2^{l+1}} = 2^{{2^l}^2} = (2^{2^l})^2$
So, if $2^{2^l} + 1$ is a prime, let p; then need prove $(p-1)^2+1$ is also.
i.e., need prove $p^2-2p+2$ is a prime, given $p$ is.

We have:
$$2^{ab} + 1 = (2^{a} + 1)(2^{a(b-1)} - 2^{a(b-2)} + \cdots + 1)$$
where $b$ is an odd integer. It's obvious that both of the factors are bigger than 1 and so if $n$ has an odd factor we $2^n + 1$ has a nontrivial factor.
If you're more interested in primes of that form, they are called Fermat's Primes and till now only five of them are known.
