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?
Your derivation (or rather, your heuristic, since there are issues of convergence of the limiting process to deal with) only gives necessary conditions for primes to be Fermat primes, not sufficient conditions. Therefore the expression you derive is simply an upper bound, not an asymptotic formula.