Let $R$ be an arbitrary local ring and $\mathfrak{m}$ its maximal ideal. Let $x$ and $y$ be in $R$ and suppose there exists a positive integer $n$ such that $xy \in \mathfrak{m}^n$ but $xy \not\in \mathfrak{m}^{n+1}$. In this case, it's not hard to see that there must exist non-negative integers $p$ and $q$ such that $x \in \mathfrak{m}^p\setminus\mathfrak{m}^{p+1}$ and $y \in \mathfrak{m}^q \setminus \mathfrak{m}^{q+1}$.
Clearly we have $n \ge p + q$. The question is, does $n = p + q$?
It's true if $\mathfrak{m}$ is principal (see below). I do not know if it's true in a general setting. Any insights, counterexamples, proofs would be appreciated! I would also be fine with a proof under some extra hypotheses on $(R, \mathfrak{m})$ like e.g. $R$ is uniserial (its ideals are totally ordered), $R$ is Noetherian (in which case $\bigcap_{k \ge 0} \mathfrak{m}^k = 0$), or at least $\mathfrak{m}$ is finitely generated, or $R$ has finite Krull dimension, etc. However, I would like to refrain from assuming $R$ is an integral domain: the result holds in $\mathbf{Z}/p^k$, for example.
Attempts. It's true if $\mathfrak{m}$ is principal: if $\mathfrak{m} = (t)$ then $\mathfrak{m}^k = (t^k)$ for all $k \ge 0$ so $x$ and $y$ are unit multiples of $t^p$ and $t^q$ respectively, whence for some unit $u$, we have $xy = ut^{p+q} \not\in \mathfrak{m}^{p+q+1}$.
It's true if $n = 1$: if $xy \in \mathfrak{m}$ then $x \in \mathfrak{m}$ or $y \in \mathfrak{m}$ (because $\mathfrak{m}$ is prime) which means that one of $p$ and $q$ is at least $1$, so $1 = n \ge p + q \ge 0 + 1 = 1$ and we have equality across the board.
I've also proved it if, say, $q = 0$. For then $y \in R \setminus \mathfrak{m} = R^\times$, so $y$ is a unit, so $yz = 1$ for some $z$ in $R$, and then $xy \in \mathfrak{m}^n$ implies $x = xyz \in \mathfrak{m}^n$ because $\mathfrak{m}^n$ is an ideal; so $p \ge n$.
Perhaps one could consider the abelian groups $V_k = \mathfrak{m}^k/\mathfrak{m}^{k+1}$: these are vector spaces over the residue field $\kappa = R/\mathfrak{m}$ (I think). The hypothesis on $y$ states that $\bar{y}$ is not zero in $V_q$. Can we say anything about the linear map $V_p \to V_n$ defined by $v \mapsto \bar{y}v$?
This is false in general. For instance, let $k$ be a field and $R=k[[t^2,t^3]]$. Then for $x=y=t^3$, we have $p=q=1$ but $n=3$, since $xy=(t^2)^3$.
If you removed the requirement that $xy\not\in\bigcap\mathfrak{m}^i$ (i.e., that $n$ exists), then your condition would be equivalent to saying that the associated graded ring of $R$ (with respect to the $\mathfrak{m}$-adic filtration) is a domain. In particular, your condition holds whenever the associated graded ring is a domain. For instance, this is true if $R$ is regular, since then its associated graded ring is a polynomial ring. If you assume $R$ is a Noetherian domain, then $xy\not\in\bigcap\mathfrak{m}^i$ always holds if $x,y\neq 0$, so your condition is equivalent to the associated graded ring being a domain.