Here is an example of a way to use $\pi$ to prove the infinitude of primes without calculating its value, or using the relatively deep fact that $\pi$ is irrational, but starting from the knowledge of $\zeta(2)$ and $\zeta(4).$
 Suppose that there were only finitely many prime numbers $ 2= p_{1}, 3= p_{2}, \ldots, p_{k-1},p_{k}.$ From the formulae $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$
and $\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}$, we may conclude after the fashion of Euler that (respectively) we have: $\prod_{j=1}^{k} \frac{p_{j}^{2}}{p_{j}^{2}-1}
= \frac{\pi^{2}}{6}$' and $\prod_{j=1}^{k} \frac{p_{j}^{4}}{p_{j}^{4}-1}
= \frac{\pi^{4}}{90}.$ Squaring the first equation and dividing by the second leads quickly to
$\prod_{j=1}^{k} \frac{p_{j}^{2}+1}{p_{j}^{2}-1}
= \frac{5}{2}$, so $5\prod_{j=1}^{k} (p_{j}^{2}-1) = 2 \prod_{j=1}^{k}(p_{j}^{2}+1).$
This is a contradiction, since the product on the left is certainly divisible by $3$, whereas every term in the rightmost product except that for $j = 2$ is congruent to $-1$ (mod 3), so we obtain $0 \equiv (-1)^{k}$ (mod 3), which is absurd.
 (I would be grateful if anyone knows a reference for a proof like this. I can't believe that I am the first person to think of it).