Noetherian module implies Noetherian ring?

I know that a finitely generated $R$-module $M$ over a Noetherian ring $R$ is Noetherian. I wonder about the converse. I believe it has to be false and I am looking for counterexamples.

Also I wonder if $M$ Noetherian imply that $R$ is Noetherian is true? And if $M$ Noetherian implies $M$ finitely generated is true?

That is, do both implications fail or only one of them?

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.

I assume that $R$ is commutative in your question. As other answers have already pointed out, the answer to your question is false for trivial reasons. But here is a closely related statement which is true. (It relies on the fact that a direct sum of two Noetherian modules is again Noetherian; I leave this for you to investigate, but feel free to leave a comment if you would like more information.)

Theorem: If a ring $R$ has a faithful Noetherian module $M$, then $R$ is Noetherian.

Proof: Because $M$ is Noetherian, it is finitely generated. Say $M = \sum Rm_i$ for some finite generating set $m_1, \dots, m_n$. Consider the homomorphism $f \colon R \to M^n$ given by $f(r) = (rm_1, \dots, rm_n)$. This is injective because the annihilator of $M$ is trivial: if $f(r) = 0$ then $rm_i = 0$ for all $i$, whence $rM = r \sum Rm_i = \sum Rrm_i = 0$. Thus $R$ is isomorphic as an $R$-module to a submodule of the Noetherian module $M^n$. It follows that $R$ is Noetherian.

Corollary: If $M$ is a Noetherian $R$-module, then $R/\mathrm{ann}(M)$ is Noetherian.

What happens for noncommutative rings? The results above no longer hold. A ring may have a faithful simple (hence Noetherian) left module but fail to be either left or right Noetherian. For instance, let $V$ be an infinite dimensional vector space over a field $k$, and let $R = \mathrm{End}_k(V)$ be the ring of $k$-linear endomorphisms of $V$. Then $V$ is a simple left $R$-module, but it can be shown that $R$ is neither left nor right Noetherian.

What about taking $R$ any non-Noetherian ring and $M=\{0\}$?

Let $R$ be a commutative non-Noetherian and let $\mathcal m$ be a maximal ideal. Then $R/\mathcal m$ is finitely generated and Noetherian - it only has two sub-$R$-modules.

Note that, even if $R$ isn't Noetherian, it contains a maximal ideal, by Krull's Theorem.