= Solution
Suppose first that $M$ is a <Noetherian module>. An ascending chain of <submodules> of $N$ is also a chain in $M$, so it stabilizes. A chain in the <quotient module> $M/N$ lifts to a chain in $M$, so it also stabilizes. Thus both $N$ and $M/N$ are <Noetherian modules>.
Conversely, let $M_1\subseteq M_2\subseteq\cdots$ be a chain of <submodules> of $M$. If $N$ and $M/N$ are <Noetherian modules>, both $M_i\cap N$ and $(M_i+N)/N$ stabilize, say for $i\geq i_0$. For $x\in M_{i+1}$, equality of the images gives $y\in M_i$ with $x-y\in N$. Then $x-y\in M_{i+1}\cap N=M_i\cap N$, and hence $x\in M_i$. This proves the extension direction of <Noetherian modules in a short exact sequence>.
The left regular <module> $R$ is <Noetherian> because $R$ is a <left Noetherian ring>. Applying the extension result inductively makes each finite <direct sum> $R^s$ <Noetherian>. Every <finitely generated module> on the left is a <quotient module> of such a <direct sum>, so it is <Noetherian>.
A <poly-(cyclic or finite) group> has a finite <subnormal series>
$$
1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_\ell=G
$$
whose factors are <cyclic groups> or <finite groups>. We prove that the <group ring> $R[G_i]$ is a <left Noetherian ring> by induction. Put $A=R[G_{i-1}]$. A finite factor makes $R[G_i]$ a finite free left $A$-<module>, using a coset transversal. Its <left ideals> are $A$-<submodules>, so <finite-module extension preserves Noetherianity> handles this case.
If the factor is infinite cyclic, choose a lift $t$ of a generator and let $\sigma(a)=tat^{-1}$. The <group ring> is the <skew Laurent polynomial ring> $A[t,t^{-1};\sigma]$. Here is the required left-sided <skew Hilbert basis theorem>. For a <left ideal> $L\subseteq A[t;\sigma]$, define
$$
I_n=\left\{\sigma^{-n}(a_n):\sum_{j=0}^n a_jt^j\in L\right\}.
$$
Each $I_n$ is a <left ideal> of $A$. Left multiplication by $t$ gives $I_n\subseteq I_{n+1}$. The <ascending chain condition> makes these ideals constant for $n\geq n_0$; choose finitely many generators for each $I_n$, $0\leq n\leq n_0$, and lift them to polynomials $f_{nj}\in L$.
If $f\in L$ has degree $d\geq n_0$ and leading coefficient $a_d$, write
$$
\sigma^{-d}(a_d)=\sum_j r_j\sigma^{-n_0}(a_{n_0j}).
$$
Subtracting $\sum_j\sigma^d(r_j)t^{d-n_0}f_{n_0j}$ cancels the leading coefficient. For $d<n_0$, use the lifts for $I_d$ instead. Induction on <polynomial degree> shows that the finitely many $f_{nj}$ generate $L$. Thus the <skew polynomial ring of an automorphism> is a <left Noetherian ring>.
Finally, for a <left ideal> $K$ of $A[t,t^{-1};\sigma]$, its intersection with $A[t;\sigma]$ has finitely many generators. Every $f\in K$ satisfies $t^mf\in A[t;\sigma]$ for some $m\geq0$, so those same generators generate $K$ after multiplication by $t^{-m}$. This completes the induction and proves that \b[$R[G]$ is left Noetherian], the left-sided form of <group rings of poly-(cyclic or finite) groups are Noetherian>.
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