Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i/M_{i-1}$ is cyclic for each $i$.

Certainly each composition is Artinian since subquotients of Artinian modules are Artinian. Also one of the composition factors must be non-Noetherian since extensions of Noetherian modules by Noetherian modules are Noetherian. Thus, we may assume that $M$ is a cyclic $R$-module.

Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say.

If $R$ is non-commutative then I'm not sure what the answer is.

Added: It seems from the wikipedia article linked from the question that Hartley showed that there are cyclic Artinian and non-Noetherian modules over certain non-commutative rings and Cohn gave another construction nearly twenty years later. See the links I give in the comments on the question for precise references.

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Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i/M_{i-1}$ is cyclic for each $i$.

Certainly each composition is Artinian since subquotients of Artinian modules are Artinian. Also one of the composition factors must be non-Noetherian since extensions of Noetherian modules by Noetherian modules are Noetherian. Thus, we may assume that $M$ is a cyclic $R$-module.

Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say.

If $R$ is non-commutative then I'm not sure what the answer is.