Are there infinitely many $x$ for which $\pi(x) \mid x$?

Well, it's not hard to show that for every natural $k>2$ equation $x=k\pi(x)$ has a positive solution.

Proof by contradiction: Imagine, that $x\ne k\pi(x)$ for every natural $x$. But then - for $x=2$:$x-k\pi(x)=2-k<0$ and for very large $x$: $x-k\pi(x)\sim x(1-\frac{k}{\ln x})>0$.

So there should be such $t$, that $t-k\pi(t)<0$ and $(t+1)-k\pi(t+1)>0$. But then from one hand:

$$ t+1-k\pi(t+1)-(t-k\pi(t))\ge 2 $$ as a difference of positive integer and negative integer, but from the other hand:

$$ t+1-k\pi(t+1)-(t-k\pi(t))=1-k(\pi(t+1)-\pi(t))\le 1 $$

Contradiction.

This is Sloane's A057809, but unfortunately there's not much information there.

Heuristically, the chance that $\pi(x)\mid x$ is $\displaystyle\frac{\log{x}}{x}$ which suggests that there are $0.5\log^2 x$ such numbers up to $x$, that is, infinitely many.