It was only last week that I noticed the awesome proof for the infinity of primes, inferred by the Fermat numbers $F_m=2^{2^m} +1,\,m=0,1,2,3,\ldots\:$ being pairwise coprime$-$a book proof indeed!
Fermat numbers satisfy the recursion $F_{m+1}=\prod\limits_{k=0}^m F_k+2$, yielding directly their coprime property.
My initial question was if every odd prime appears as a divisor of some Fermat number?!
After some enquiry I found Édouard Lucas' Theorem saying that a prime divisor $p$ of $F_m$ where $m\geqslant 2$ has the form $\,p=2^{m+2}k+1\,$ for some $k$, or equivalently $\,2^{m+2}\mid (p-1)\,$. This excludes primes
$p\equiv 3\pmod 4$, with the exception $3=F_0$.
Thus focusing on prime divisors $p\equiv 1\pmod 4$ lets me ask for a more concise description:
Can one characterise in a (preferably) uniform way all those primes $\equiv 1\pmod 4\,$ which do not divide any Fermat number?
The comments received until now hint at "No" as answer, or more reservedly, that this is not known.