It is not clear to me exactly what you are asking in your main question. If you are asking:
$\bullet$ If a ring $R$ admits a Noetherian module, must $R$ be Noetherian?
Then this is trivially false: Alex Youcis and Thomas Andrews have each shown that every commutative ring admits Noetherian modules. (As a general rule, if you are looking for a counterexample to an assertion about modules and you haven't checked the zero module, you haven't looked hard enough. Also looking at modules of the form $R/I$ is something to try early on.)
If you are asking
$\bullet$ If for a ring $R$ every finitely generated $R$-module is Noetherian, must $R$ be Noetherian?
Then this is trivially true, as $R$ is a finitely generated $R$-module.
A less trivial statement is the following:
Lemma (Kaplansky): A ring is Noetherian iff it admits a faithful Noetherian module.
Another result vaguely along these lines is:
Theorem (Eakin-Nagata) Let $R \subset S$ be a ring extension such that $S$ is finitely generated as an $R$-module. Then $R$ is Noetherian iff $S$ is Noetherian.
Proofs of these and other results which are (even more) vaguely related to your question can be found in $\S 8.8$ of my commutative algebra notes.