Suppose $R$ is a commutative ring (but see the edit portion below) and $\mathfrak{m}$ is a maximal ideal of $R$ such that $|R/\mathfrak{m}|<\infty$. Also assume that $k$ a positive integer. Is it true that for a prime number $p$, we have $p$ dividing $|R/\mathfrak{m}^k|$ if and only if $p$ is the characteristic of the field $R/\mathfrak{m}$?
Attempt: One direction is clear because I know (by $0\to\mathfrak{m}^{i-1}/\mathfrak{m}^i\to R/\mathfrak{m}^i\to R/\mathfrak{m}^{i-1}$ repeatedly) that
$$|R/\mathfrak{m}^k|=|R/\mathfrak{m}|\cdot|\mathfrak{m}/\mathfrak{m}^2|\dots|\mathfrak{m}^{k-1}/\mathfrak{m}^k|$$ so if $p$ is the characteristic of $R/\mathfrak{m}$, then $p$ divides $|R/\mathfrak{m}|$ (the additive order of $1$ is $p$ which divides the order of the additive group) and hence divides $|R/\mathfrak{m}^k|$.
However, I am not too sure about the other direction. Specifically, it could be that $p$ divides $|R/\mathfrak{m}^k|$ but the fatcor of $p$ shows up in the $|\mathfrak{m}^{i-1}/\mathfrak{m}^i|$ factors rather than $|R/\mathfrak{m}|$.
Edit: I guess I should be clear that in my original context for this question, $R$ is actually the ring of integers $O_K$ for some number field $K$ (finite extension of $\mathbb{Q}$). This might affect the analysis (e.g. so $R$ is a Dedekind domain, and also we know $|R/I|<\infty$ for all ideals $I$).
Start with $$|R/\mathfrak{m}^k|=|R/\mathfrak{m}|\cdot|\mathfrak{m}/\mathfrak{m}^2|\cdots|\mathfrak{m}^{k-1}/\mathfrak{m}^k|$$ and use that $\mathfrak m^{i-1}/\mathfrak m^i$ is an $R/\mathfrak m$-vector space, so isomorphic to a direct sum of copies of $R/\mathfrak m$.