For a ring $R$ with maximal ideal $I$, is there an extension for powers of $I$ like for $n \in \mathbb{N}$ such as $$ rs \in I^n, s \not \in I \implies r \in I^n $$
I was at first looking for general prime ideals, but Not primary ideal having a prime radical which yields a prime counter-example, but the comments on the question mention maximal ideals being okay, hence why I edited the question.
Under the same assumption we clearly have $r \in I$, but I'm stumped on this extension. It clearly holds true in $\mathbb{Z}$ by factoring the elements into primes, but in a general ring we might not have a factorisation into irreducibles, so I'm wondering whether it's possible to show/disprove this.
My current thoughts are that quotienting by $I^n$, $$ (r+I^n)(s+I^n) = 0 + I^n $$ but we can't then say one of the LHS terms must be zero since $R/I^n$ isn't necessarily an integral domain... Which makes me think there's a possibly very simple counterexample that I'm missing, or a proof which has a few more details.
This is true; from the facts that $s \notin I$ and that $I$ is maximal, you can deduce the existence of some $t \in R$ with $st + 1 \in I$. By raising this expression to the power $n$, you can verify that this leads to some element $t' \in R$ such that $st' + 1 \in I^n$. Hence $r(st' + 1) = rst' + r \in I^n$. Since $rs \in I^n$, it follows from this that $r \in I^n$.