Proof
First, we prove a lemma: Let $a$ be a positive integer greater than 1, and $q$ be an odd prime factor of the number $a^{2^{n}}+1$, then
$$q \equiv 1\left(\bmod 2^{n+1}\right)$$

In fact, from $a^{2^{n}} \equiv-1(\bmod q)$, we know
$$a^{2^{n+1}} \equiv 1(\bmod q),$$

This indicates that $\delta_{q}(a) \nmid 2^{n}$, but $\delta_{q}(a) \mid 2^{n+1}$. Therefore, $\delta_{q}(a)=2^{n+1}$ (note that we have used $q>2$ here). Then, by Fermat's Little Theorem, we know $a^{q-1} \equiv 1(\bmod q)$, hence $2^{n+1} \mid(q-1)$, i.e.,
$$q \equiv 1\left(\bmod 2^{n+1}\right)$$

The lemma is proved.
Returning to the original problem. When $n>1$, note that,
$$\begin{aligned} F_{n-1}^{2^{n+1}} & =\left(2^{2^{n-1}}+1\right)^{2^{n+1}}=\left(2^{2^{n}}+2^{1+2^{n-1}}+1\right)^{2^{n}} \equiv\left(2^{1+2^{n-1}}\right)^{2^{n}} \\ & =\left(2^{2^{n}}\right)^{1+2^{n-1}} \equiv(-1)^{1+2^{n-1}} \\ & \equiv-1\left(\bmod F_{n}\right) \end{aligned}$$

Therefore, for a prime factor $q$ of $F_{n}$, we have $q \mid\left(F_{n-1}^{2^{2+1}}+1\right)$. Using the conclusion of the lemma, we get
$$q \equiv 1\left(\bmod 2^{n+2}\right)$$