I have been having problems finding a solution for this problem and honestly have no ideas left how to solve this, please help.
Assume that $b^n + 1= $ a prime number for some integers $b,n$ where $b>1$ and $n>1$. Prove that $n$ must take the form $n=2^k$ for some positive integers $k$
What i have been looking at are the generalized Fermat numbers due to $b^n + 1$ taking the form ${b^2}^k + 1$ for $n=2^k$ and Fermat's little theorem but seemingly without any progress.
If $n$ is an odd natural number then $-1$ is a root of the polynomial $P(x)=x^n+1$ and hence $x+1|x^n+1$.
Now suppose $b^n+1$ is prime and using prime decomposition write $n=2^rm$ when $m$ is odd. Then $b^n+1=(b^{2^r})^m+1$ and hence $b^{2^r}+1|b^n+1$. But we supposed $b^n+1$ is prime so from here we conclude that $b^{2^r}+1=b^n+1$. This implies $n=2^r$.
Suppose that $b^n+1$ is prime, $b>1$ and $n>1$. We shall show that $n$ has no odd factor except for $1$: this proves that $n$ is a power of $2$.
So, let $n=st$ where $s,t$ are positive integers and $s$ is odd. You should know the factorisation $$x^s+1=(x+1)(x^{s-1}-x^{s-2}+\cdots-x+1)\ .$$ Substituting $x=b^t$ shows that $b^t+1$ is a factor of $b^n+1$. But $b^n+1$ is prime, so there are two options: $$b^t+1=1\quad\hbox{or}\quad b^t+1=b^n+1\ .$$ The first is clearly impossible since $b>1$; and the second gives $t=n$, hence $s=1$. We have shown that that only possible odd factor of $n$ is $1$, and as explained above, this means that $n$ is a power of $2$.
