I have proven in the past that if $A$ is a local ring with maximal ideal $\mathfrak{m}$ so that $\mathfrak{m}$ is principal, and $\bigcap_{n>0}\mathfrak{m}^n = \{0\}$, then $A$ is necessarily Noetherian, since every ideal is a power of $\frak{m}$.
Conversely, the Krull Intersection Theorem guarantees that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$, then $\bigcap_{n>1}\mathfrak{m}^n = \{0\}$.
I'm wondering about the first theorem without the assumption that $\mathfrak{m}$ is principal, and is instead just finitely generated? Here is a demonstration of where my previous proof breaks down:
Let $x \in I$ for some ideal of $A$. Assuming (wlog) that $I$ is not maximal, we know that $I \subseteq \mathfrak{m}$. Choose $n$ so that it is the maximal power where $I \subseteq \mathfrak{m}^n$ since it would otherwise be trivial, by assumption.
If $\mathfrak{m}$ were principal, then $x = a \cdot m^n$, where $m$ is the generator of $\mathfrak{m}$. When $\mathfrak{m}$ is principal, I could then argue that $I = \mathfrak{m}^n$ since $a$ would then be a unit. However, I don't think this argument works here.
If this does not work, I was wondering if maybe using the fact that the natural homomorphism
\begin{equation} \phi\colon A \to \prod_{i>0}A/I_i \end{equation}
is an injection, since $\ker(\phi) = \bigcap I_i = \{0\}$, but I'm unsure of how I can use this observation.
So: is the assumption that the maximal ideal is finitely generated sufficient here? If yes, is there any way to recover either of the "partial" proof ideas that I was concerned with?
So it's not true that a local ring with finitely generated maximal ideal whose intersection of powers is zero has to be Noetherian. You can even have an example where the maximal ideal is generated by 2 elements but it's not that easy to find. One explicit such ring can be found in this paper.
The ring $B$ in Proposition 3.2 on page 6 is local with $2$-generated maximal ideal. Note 3.5(2) on page 7 says that the intersection of powers of the maximal ideal is zero. $B$ is not Noetherian because $\dim B = 3$ (so if $B$ were Noetherian then Krull's Principal Ideal Theorem would imply the maximal ideal needs at least $3$ generators).