I am supposed to prove: $$\underbrace{\pi^2 \frac{\cos(\pi z)}{\sin^{2}(\pi z)}}_\text{:=f(z)}=\underbrace{\sum_{n \in \mathbb{Z}} \frac{(-1)^n}{(z-n)^2}}_\text{:=g(z)}$$
I was able to show that both sides have the same poles and the same singular parts. Therefore, $h=f-g$ is an entire function.
Clearly, both $f$ and $g$ have period $2$, so $h$ also has period $2$. So, if I can show that $h(z)$ is bounded in the vertical strip $0 \leq \Re(z) \leq 2$, I can invoke Liouville's to get $h$ is consant.
To evaluate that constant, one can use the fact that $\lim_{z \to 0}\Big(f(z)-\frac{1}{z^2}\Big)=-\frac{\pi^2}{6}=\sum_{n\neq 0} \frac{(-1)^n}{(z-n)^2}$ implying $h \equiv 0$ and we are done.
So, how should I show that $h$ is bounded in that strip?
Note first that since $h$ is entire it is bounded in the rectangle $0\le x\le 2$, $|y|\le1$. So it's enough to show that the sum on the right is bounded in the set $0\le x\le 2$, $|y|>1$. We assume $|y|>1$ below.
If $z=x+iy$ then $$\sum_n\frac1{|z-n|^2}=\sum_n\frac1{(x-n)^2+y^2}.$$The three or five terms where $n$ is closest to $x$ are each bounded by $1/y^2<1$. For the other values of $n$ you actually have $|x-n|\ge c|N-n|$ where $N$ is the integer closest to $x$, so those terms are bounded by $1/(c(n-N))^2$. (It's easy to see $N\ne n$.)
Details: Fix $x$. There are only five or so values of $n$ with $|x-n|\le2$. Suppose $|x-n|>2$. Note that $n\ne N$, since $|x-N|\le1/2$; hence $|N-n|\ge1$. Now $$|x-n|=|(x-N)+(N-n)|\ge|N-n|-|x-N|\ge|N-n|-\frac12\ge|N-n|-\frac12|N-n|=\frac12|N-n|.$$