Does anyone know if there is a value of $z$ between $0$ and $\pi$, such that the product below equals zero?
$$ \lim_{m \rightarrow \infty} \prod_{k=1}^{m}\left(1-\frac{z^2}{4k^2\pi^2}\right) \prod_{k=1}^{m}\left(1-\frac{z^2}{9k^2\pi^2}\right)...\prod_{k=1}^{m}\left(1-\frac{z^2}{n^2k^2\pi^2}\right)$$
Set $s = \sqrt \frac {z^2} {\pi^2}$. Then you're asking if series $$\sum_{k=1}^{\infty} \sum_{l = 2}^{n} log(1 - \frac {s} {(lk)^2})$$ diverge for $s \in (0, 1)$.
All terms are negative, so convergence does not depend on order; change summation order and take $l = 2$ part — if it converges, other do too. If $k > 1000$, then we may estimate $$|log(1 - \frac {s} {4k^2})| < \frac {1} {k^2} $$ and therefore your product is always nonzero.