Euler's proof that the fifth Fermat number is composite begins with the following argument. If $p$ divides $F_n$, then the order of $2$ in $(\mathbb Z/p\mathbb Z)^*$ is exactly $2^{n+1}$. Hence $p\equiv1$ mod $2^{n+1}$. If $n\ge2$, this implies that $2$ is a square mod $p$, $2=\omega^2$. Then the order of $\omega$ is exactly $2^{n+2}$, hence $p\equiv1$ mod $2^{n+2}$. This limits the set of prime divisors to be tested. For $n=5$, we find $p=257, 641, ...$, in which the first one is impossible because it is $F_3$ and the Fermat numbers are pairwise coprime. Thus the first candidate is $641$, which turns out to divide $F_5$.
Of course, the situation is not so simple for larger values of $n$. But I wander how far this argument could be pushed forward. Because a prime divisor satisfies $p\equiv1$ mod $2^{n+2}$, $2$ should be a $4$-th power if and only if $2^{\frac{p-1}{4}}\equiv1$ mod $p$. If $2=\theta^4$, then ...(bla-bla)... $p\equiv1$ mod $2^{n+3}$. So my question is
Let $m\ge2$ be given. Assume that $p-1$ is divisible by a large enough power of $2$. Does this imply that $2$ is a $2^m$-th power mod $p$ ?
Of course, I am not so naive as to look forward for an elementary solution to the long-standing problem of the primality of some $F_n$'s, but such an approach can provide nice exercises in undergraduate classes.