Here's the conjecture:
For $n\in\mathbb{N}^*$, $(2^n+1)$ always has a prime factor with form $(2nk+1)$ where $k\in\mathbb{N}^*$, with an exception when $n=3$.
For example, when $n=5$, we have $2^5+1=33=3\times 11$, here the factor $11$ can be written as $2\times 5\times 1+1$, which means $k=1$ satisfies the conjecture.
I've verified this conjecture for $n\le 100$ but can't come up with an idea to prove or disprove it. I noticed that Zsigmondy's Theorem has a similar form but I can't find a good way to apply it in this problem. Any help would be appreciated!
By Zsigmondy's theorem there is an odd prime $p$ which divides $2^n+1$ but does not divide any number of the form $2^k+1$ for $k< n$. We have $2^n\equiv -1\pmod{p}$, so $2^{2n}\equiv 1\pmod{p}$. If you manage to prove that $2n$ is exactly the order of $2$ in $\mathbb{Z}/p\mathbb{Z}^*$ you are done, since the order of an element has to divide the order of the group, and $2n\mid (p-1)$ is equivalent to $p=2nk+1$. Essentially you just have to prove that $2^d\not\equiv 1\pmod{p}$ for any divisor $d$ of $2n$.