I have a sneak feeling that for $n\geq2$ we have that $2^{2^n}+1$ is a factor of $2^{2^{2^{n}-1}}-1$. If we say that $2^{2^{2^{n}-1}}=2^{2^n}\cdot2^k$ for some positive integer $k$ then my idea was to write $$2^{2^{2^{n}-1}}-1=(2^{2^n}+1)(2^k+\dots-1)$$ but I'm having trouble finding the explicit form this second bracket on the right hand side.
I'm welcome to any ideas.
The general result is that $(a-b)|(a^m-b^m)$ which you can prove by summing a geometric series with ratio $\frac ab$. Note that $\left(2^{2^n}+1\right)\left(2^{2^n}-1\right)=2^{2^{n+1}}-1$ so apply the general result with $a=2^{2^{n+1}}, b=1, m=\frac {2^{2^{n-1}}}{2^{n+1}}$. $m$ will be an integer if $2^{n-1} \ge n+1$, which is true for $n \ge 3$. I have checked the case $n=2$ separately, giving $17|255$.