Let every prime of the form $2^n+1$ be called "Fermat Prime" (I know that the real definition is by using $2^{2^{n}}+1$, but I will use the other one to get things easier). By definition, we have that $p$ will be a Fermat Prime if and only if it is prime and it is not of the form $qn+1$ for any prime $q$ such that $2<q<p$.
Then, by Dirichlet Theorem on arithmetic progressions, we have that, as $N \to \infty$, the number of primes of the form $qn+1$ lower than $N$ will tend to be the same than the one of primes of the form $qn+2$ and of the one of the primes of the form $qn+3$; and like this up to $qn+(q-1)$.
So, to know the number of Fermat Primes as $N \to \infty$, we have that:
Only the primes of the form $3n+2$ can be a Fermat Prime. Then, as $N \to \infty$, The number of FM will be lower than $\pi(N)\frac{1}{2}$.
Only the primes of the form $5n+2, 5n+3$ and $5n+4$ can be a Fermat Prime. Then, as $N \to \infty$, The number of FM will be lower than $\pi(N)\frac{1}{2}\frac{3}{4}$.
...
If we continue like this, we would get that: $$\lim_{N \to \infty}\pi_{\text{FermatPrimes}}(N)=\lim_{N \to \infty} \pi(N)\prod_{p>2}\frac{p-2}{p-1}=K \lim_{N \to \infty} \frac{\pi(N)}{\log{(N)}}$$
Which clearly diverges.
Is this true?