Because any divisor of $F_n$ is a product of some prime divisors of $F_n$, therefore it is sufficient to show that any prime divisor of $F_n$ is congruent to $1$ modulo $2^{n+1}$.

Let $p$ be any prime divisor of $F_n$. It is obvious that $p\ne 2$. Then, because $p\mid F_n=2^{2^n}+1$, we have $$2^{2^n}\equiv -1\pmod{p} \implies 2^{2^{n+1}}\equiv 1\pmod {p}$$

Let $r$ be the order of $2$ modulo $p$. Then $r\mid 2^{n+1}$ which means $r$ is some powers of $2$. Let $r=2^m$ where $0\le m\le n+1$.

If $m \le n$, then $2^{2^m}\equiv 1\pmod{p}$. Continuously taking square of this relation will eventually yield $2^{2^n}\equiv 1\pmod{p}$. Considering this together with the first equation above, i.e. $2^{2n}\equiv -1\pmod{p}$, will force $p=2$. This contradicts to the fact that $p\ne 2$. Hence, we conclude $m$ has to equal $(n+1)$, or $r=2^m=2^{n+1}$.

Meanwhile, by Fermat's little theorem, we have $2^{p-1}\equiv 1\pmod{p}$. Therefore $$r\mid (p-1) \implies 2^{n+1}\mid (p-1)\implies p\equiv 1\pmod{2^{n+1}}$$