Suppose $(R,m)$ is a local ring with $m = (x)$ a principal ideal. Suppose that $x^{k} \ne 0 , \forall k \geq 0$. Do we have that $I = \cap_{k \in \mathbb{N}} (x^{k})$ is a finitely generated ideal, and if so why?
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In general, no. Consider the ring of germs at $0$ of $C^{\infty}$ functions from $\mathbb{R}$ to $\mathbb{R}$. This is a local ring with maximal ideal $(x)$, but we have $e^{-1/x^m}\in (x^n)$ for all $n\geq 1$ and for all $m\geq 2$ and these functions are suitably independent.
If $R$ is Noetherian, then not only is $\cap_{n>1} (x^n)$ finitely generated, it's equal to $0$! In fact, this condition is sufficient - a local ring $(R,\mathfrak{m})$ with principal maximal ideal $\mathfrak{m}$ is Noetherian if and only if $\cap_{n>1} \mathfrak{m}^n=0$.
