How does the probability of a number being prime increase, if its factors have the form $km+1$? The "probability" that a number $n$ is prime, is roughly $\frac{1}{\ln(n)}$.

How does this probability increase, if it is known, that the prime factors must
   have the form $km+1$ for some given $m\in \mathbb N$ ?

Motivation : I wonder, how likely the generalized fermat numbers $b^{2^l}+1\ ,\ b\ even\ $
are prime. It is known that the factors of such a number must have the form
$2^{l+1}k+1$. This should increase the probability of being prime, but I
have no idea of the magnitude of the increase.
 A: Let $S_m$ be the set of integers whose prime factors are all congruent to $1$ mod $m$.  We know from the Prime Number Theorem for Arithmetic Progressions, as described in Julián's answer, that
$$ \# \{ p \le x : p \in S_m, p \text{ prime} \} = c_m \frac{x}{\log x} + O_m\big(\frac{x}{\log^2 x}\big),$$
where $c_m = 1/\phi(m)$.  To answer the question, we only need an asymptotic for the counting function of $S_m$.  Since $S_m$ excludes a fraction of $(1-c_m)$ of all primes, one expects heuristically that
$$ \# \{ n \le x : n \in S_m \} \asymp \frac{x}{(\log x)^{1-c_m}},$$
and in fact it can be obtained rigorously by a theorem of Odoni that the following precise asymptotic holds:
$$ \# \{ n \le x : n \in S_m \} = D_m x(\log x)^{c_m-1} + O_m(x(\log x)^{c_m-2+\epsilon}),$$
where $D_m > 0$ is given by a certain convergent Euler product.  Specifically, if $\chi_m(p)$ is the indicator function of $\{ p : p \equiv 1 \pmod m\}$, then
$$D_m = \frac{1}{\Gamma(c_m)} \prod_{p} \big(1 - \frac1p\big)^{c_m-\chi_m(p)}.$$
The "probability" you're seeking is just the ratio of primes to all numbers in $S_m$ up to $x$, namely $$\frac{c_m}{D_m (\log x)^{c_m}}.$$
I think in this case, the $\epsilon$ in the error term can be removed by applying the Selberg-Delange method, but I haven't checked the details (see, for instance, this paper).
However, as pointed out in Aravind's answer, even though this is provable, it still only amounts to a heuristic when you try to apply it to much thinner sets such as $\{n^2+1\}$ or $\{2^{2^n}+1\}$ that are not defined purely in terms of the prime composition.
A: I believe this question is impossible to answer in general and very hard for even simple instances.
Firstly, we are looking at a set $S$ of numbers such that all prime factors of numbers in $S$ are of the form $km+1$. The question is about the fraction of primes in $S$ among the first $n$ numbers in $S$.
It is not clear why this function should be greater than $\frac{1}{\log n}$ or indeed how we can say anything about it for arbitrary $S$. If we choose $S$ as a subset of the composites, then the probability is zero, while we could make it 1 by choosing $S$ to be a subset of the primes.
Also consider the example $S=\{n^2+1: n \in \mathbb{N}, \text{ n even}\}$. We know that all prime factors must be of the form $4k+1$, but clearly this is not sufficient information; we don't even know if the probability is positive! 
The same is true for any set $S$ for which whether it has infinitely many primes is itself open, such as the set of Fermat numbers.
A: The probability that a number smaller than $n$ of the form $k\,m+1$ is prime is of the order of
$$
\frac{m}{\phi(m)}\,\frac{1}{\log N}\ .
$$
This is an upper bound on the probability you are looking for. To get a better estimate you would need to know the number of integers less than $n$ all whose factors are $\equiv1\mod m$.
