There is a general strategy, which unfortunately is a failure because the information needed for success is scarcely ever available.

Let $A$ denote a set of primes, and $B$ the set of all products of powers of the primes in $A$. Then

$$
\prod_{p \in A}(1 - p^{-s})^{-1} = \sum_{n \in B}n^{-s}
$$

by the Euler product formula. The Dirichlet series on the right hand side has nonnegative coefficients, so by a theorem of Landau, the point where the line of convergence crosses the real axis is a singularity of the sum function. If $A$ is finite, this singularity is at the origin. Thus if it can be established that the line of convergence is to the right of the imaginary axis, $A$ must be infinite (The only reason to invoke the theorem of Landau is to guarantee that there always is a singularity on the line of convergence). Of course, existence of a singularity of the sum function somewhere in the open right hand half plane would also work, even if we do not know the line of convergence.

This tends to fail because we can't get a good grip on $B$. Suppose $A$ is the set of twin primes (primes $p$ such that $p+2$ is also prime). Nothing really useful is known about the set $B$ of integers that are products of powers of twin primes. But if the accepted conjecture about the distribution of twin primes holds, there will be a singularity at $s = 1$, so $B$ cannot be really sparse.

As an example, there are infinitely many primes $p \equiv 1\pmod{4}$. Letting these primes, together with $2$, constitute $A$, we see that $B$ contains the
values of the polynomial $n^2 + 1$, since the latter is never divisible by any prime $q \equiv 3\pmod{4}$. Then

$$
\sum_{n \in B}n^{-\sigma} \geq \sum_{m = 1}^{\infty}(m^2 + 1)^{-\sigma}
$$

and thus the Dirichlet series over $B$ has a singularity in the half plane $\sigma \geq 1/2$. So $A$ has to be infinite.

Admittedly, this example is not that interesting.