So I was looking at the Fermat Primes. These are primes of the form $2^k+1$ for a natural number $k$, such that I define by $\mathbb{N}:=\big\{1,2,3,\ldots\big\}$ and $0\notin \mathbb{N}$. We denote by $F_1$ the first Fermat Prime; $F_2$ the second Fermat Prime; et cetera up to $F_k$.
Thus far, however, only $5$ Fermat Primes are known. They are the following, respectively: $$3,5,17,257,65537,\ldots$$ I write $(\ldots)$ because I assume that there exist more Fermat Primes, but it is a conjecture that the above Fermat primes are the only ones.
I introduce this to you in the event that you do not know of them, because my question very much relates to these Fermat Primes.
If $2^k+1$ is prime, then must it be true that $k$ is strictly a power of $2$?
I noticed that, $$\begin{align}3&=2^{2^0}+1\\ 5&=2^{2^1}+1 \\ 17&=2^{2^2}+1 \\ 257&=2^{2^3}+1 \\ 65537&=2^{2^4}+1.\end{align}$$
Attempt at Proof:
I proved that $k$ is even, because supposing otherwise implies that $k=2j+1\,\exists j\in\mathbb{N}$ and $$2^k+1=2^{2j+1}+1=2^{2j+1}-(-1)^{2j+1}\stackrel{\centerdot}{:}2-(-1)=3.\tag*{$\bigg(\begin{align}&\text{$a\stackrel{\centerdot}{:}b$ is read as $a$} \\ &\text{is divisible by $b$.}\end{align}\bigg)$}$$ Ergo, $2^k+1$ is not prime if $k$ is odd; $k$ must be even. $\;\bigcirc$
Now if $k$ is even, then $k=2j$ and $$2^k+1=2^{2j}+1=4^j+1.$$ If $j$ is odd, then likewise, as similarly demonstrated before, $$4^j+1=4^j-(-1)^j\stackrel{\centerdot}{:}4-(-1)=5.$$ Ergo, $l\nmid k\,\exists l$ odd; $k$ must be a power of $2$ as it will only then have no odd divisors. $\;\bigcirc$
Is my proof correct? I don't see anything wrong with it, but if it is correct, are there other (better) proofs? Also, is there anything new on Fermat Primes; particularly, are there any other general constraints on $k$? How far has the conjecture been tested for? I checked the first $250$ Fermat Primes and it seems like the conjecture is true.
(If you want, you can go here, type x=2;x=2x-1;c<=250;x and then wait a minute before the first $250$ Fermat Primes appear. I still have to thank one of the users on the MSE who introduced me to the site accessible via the link.)
Thank you in advance.
A simple proof is based on the factorization of $x^n+1$ when $n$ is odd: $$ x^n+1 = (x+1)(x^{n-1}-x^{n-2}+\cdots+1) $$ Therefore, if $m=nd$ with $n$ odd, then $x^d+1$ divides $x^m+1=(x^d)^n+1$.
In particular, $2^m+1$ is divisible by $2^d+1>1$ and so is not prime.
Thus, if $2^m+1$ is prime, then $m$ has no odd factor and so is a power of $2$.