Let me take an extension ring of the localization $\Bbb{Z}_{(2)}$ of $\Bbb{Z}$ which is not a DVR: $$R=\Bbb{Z}_{(2)}\left[\sqrt{-3}\right]$$
I want to show that $R$ is a noetherian ring.
My idea was the following. We know that $R=\Bbb{Z}_{(2)}\left[\sqrt{-3}\right]=(\Bbb{Z}\setminus (2))^{-1}\Bbb{Z}\left[\sqrt{-3}\right]$. But now I know that $\Bbb{Z}\left[\sqrt{-3}\right]\cong \Bbb{Z}[X]/(X^2+3)$. Since $\Bbb{Z}[X]$ is noetherian and $(X^2+3)$ is an ideal we know by a corollary of Hilbert's basis theorem that $\Bbb{Z}\left[\sqrt{-3}\right]$ is noetherian. But then localization doesn't affect noetherianity so it is also still noetherian.
Does this work?
Thanks for your help.
Just as a point of clarification from the comments: neither of the isomorphisms $\Bbb{Z}_{(2)}\left[\sqrt{-3}\right]=(\Bbb{Z}\setminus (2))^{-1}\Bbb{Z}\left[\sqrt{-3}\right]$ nor $\Bbb{Z}\left[\sqrt{-3}\right]\cong \Bbb{Z}[X]/(X^2+3)$, nor, for that matter, the upstairs analogue $\Bbb{Z}_{(2)}\left[\sqrt{-3}\right]\cong \Bbb{Z}_{(2)}[X]/(X^2+3)$, are necessary here.
Step 1. $\mathbb Z_{(2)}$ is Noetherian. You understand this part.
Step 2. $\mathbb Z_{(2)}[X]$ is Noetherian by Hilbert's Basis Theorem.
Step 3. There is a surjection $\mathbb Z_{(2)}[X] \twoheadrightarrow \Bbb{Z}_{(2)}\left[\sqrt{-3}\right]$, defined by identity on $\mathbb Z_{(2)}$ and $X \mapsto \sqrt{-3}$. The ideals of the codomain of a surjective ring homomorphism are in one-to-one correspondence with a proper subset of the ideals of the domain (specifically, the ideals which contain the kernel of that map - no need to know what that kernel is). The proper subset of ideals automatically satisfies the ascending chain condition.