Partial answer: if $\nu$ is semifinite then it is $\sigma$-finite.
Following the setup of the first version of your statement, suppose that $\pi$ is $\sigma$-finite and satisfies $\pi(A \times B) = \mu(A) \nu(B)$ and that $\nu$ is semifinite. (We do not actually need the assumption that $\mu$ is $\sigma$-finite.) We show $\nu$ is $\sigma$-finite.
We follow the idea from Uniqueness of product measure (non $\sigma$-finite case). Since $\pi$ is $\sigma$-finite, there is a sequence of sets $E_n \subset X \times Y$ with $\pi(E_n) < \infty$ and $X \times Y = \bigcup_n E_n$. For each $n$, let
$$a_n = \sup\{ \pi((X \times B) \cap E_n) : \nu(B) < \infty\}.$$
Note that $a_n \le \pi(E_n) < \infty$. Now by definition of sup, there is a sequence of sets $B_{n,k} \subset Y$ with $\nu(B_{n,k}) < \infty$ and $\pi((X \times B_{n,k}) \cap E_n) \uparrow a_n$. Set $C = Y \setminus \bigcup_{n,k} B_{n,k}$. I claim $\nu(C) =0$, which would complete the proof.
By semifiniteness, it is enough to show that for every measurable $C' \subset C$ with $\nu(C') < \infty$, we have $\nu(C') =0$. Now since such $C'$ is disjoint from all the $B_{n,k}$, we have for every $k$ that
$$\pi((X \times B_{n,k}) \cap E_n) + \pi((X \times C') \cap E_n) = \pi((X \times (B_{n,k} \cup C')) \cap E_n) \le a_n$$
since $\nu(B_{n,k} \cup C') < \infty$. But by definition of the $B_{n,k}$, we have $\sup_k \pi((X \times B_{n,k}) \cap E_n) = a_n$, so we conclude $\pi((X \times C') \cap E_n) = 0$. Since $X \times Y$ is the countable union of the $E_n$, it follows that $0 = \pi(X \times C') = \mu(X) \nu(C')$. Since $\mu(X) > 0$ by assumption, we have $\nu(C') = 0$ as desired.
I don't know what happens if $\nu$ is not semifinite. You would think that if $\nu$ is not semifinite then it would make it even harder for $\pi$ to be $\sigma$-finite, but I don't see how to prove that at the moment.
Best Answer
Hint: Let $f: X \to \mathbb{R}$ be non-negative and in $L^2$. Separate $f = f|_{\{f\le 1\}}+f|_{\{f> 1\}}$.
For the first part, notice that the measure space is finite. For the second part, notice that for some (which ones?) $\mathbb{R}\ni a>0$ we have $a \le a^2$.