= Solution
Use unital <rings>, and distinguish left and right <Noetherianity>. A useful form of the <noncommutative Hilbert basis theorem> is the following: if a <ring> $B$ is generated by a subring $A$ and an element $x$, if $A+xA=A+Ax$, and if $A$ is a <right Noetherian ring>, then $B$ is a <right Noetherian ring>. The corresponding left-handed statement holds as well. In particular, a central <polynomial ring> over a <Noetherian ring> is <Noetherian>, and a <skew polynomial ring of an automorphism> preserves either side's <Noetherianity>.
Here is a proof of the general right-handed statement. Put $F_{-1}=0$ and
$$
F_n=A+Ax+\cdots+Ax^n=A+xA+\cdots+x^nA.
$$
The equality follows inductively from $A+xA=A+Ax$. These groups form an exhaustive <ascending filtration>, stable under multiplication by $A$ on either side, with $F_jF_k\subseteq F_{j+k}$. For any $b\in A$ and $n\geq0$, moving coefficients one step at a time also gives
$$
bx^n=x^nc+v\qquad(c\in A,\ v\in F_{n-1}).
$$
Fix a <right ideal> $I$ of $B$ and define coefficient sets
$$
L_n=\{a\in A:ax^n\in I+F_{n-1}\}.
$$
Each $L_n$ is a <right ideal> of $A$. Additive closure is immediate. If $a\in L_n$ and $b\in A$, use the preceding transport identity to obtain
$$
abx^n=(ax^n)c+av\in I+F_{n-1}.
$$
Here $(ax^n)c$ has this membership because $I$ is a <right ideal> and $F_{n-1}$ is stable under right multiplication by $A$. Also $L_n\subseteq L_{n+1}$, by multiplying $ax^n\in I+F_{n-1}$ on the right by $x$. The <ascending chain condition> in $A$ yields $N$ with $L_n=L_N$ for $n\geq N$.
For each $0\leq k\leq N$, choose finite right generators $a_{kj}$ of $L_k$ and lifts $f_{kj}\in I$ with $f_{kj}-a_{kj}x^k\in F_{k-1}$. These finitely many lifts generate $I$ as a <right ideal> of $B$. To prove this, let $f\in I\cap F_d$ and write $f=ax^d+v$, $v\in F_{d-1}$. Then $a\in L_d$. Put $k=\min(d,N)$ and write $a=\sum_ja_{kj}b_j$. Transport each coefficient as $b_jx^d=x^dc_j+v_j$ with $v_j\in F_{d-1}$. Consequently
$$
f-\sum_j f_{kj}x^{d-k}c_j\in I\cap F_{d-1}.
$$
Induction on $d$, with $F_{-1}=0$, expresses $f$ in the right ideal generated by the lifts. Every <right ideal> is therefore finitely generated, proving the theorem. Passing to the <opposite ring> proves the left-handed version. This proof does not require unique coefficient normal forms.
A <poly-(cyclic or finite) group> has a finite <subnormal series>
$$
1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_r=G
$$
whose factors $G_i/G_{i-1}$ are cyclic or finite. Normality is required between consecutive terms. We show inductively that each <group ring> $\mathbb Z[G_i]$ is both a <left Noetherian ring> and a <right Noetherian ring>.
The starting <ring> $\mathbb Z$ is <Noetherian>, since every <ideal> is principal. The other module fact we use is that a <finitely generated module> over a <Noetherian ring> on the relevant side is a <Noetherian module>: it is a quotient of a finite <direct sum> of regular modules, and the <ascending chain condition> passes to finite sums and quotients. For a finite sum this follows by induction, comparing the intersections of a chain with one summand and its images in the quotient by that summand.
If $H\triangleleft K$ has finite index, choosing coset representatives shows that $\mathbb Z[K]$ is finite free as a left and as a right $\mathbb Z[H]$-<module>. By <finite-module extension preserves Noetherianity>, every one-sided <ideal> of $\mathbb Z[K]$ is finitely generated already as a $\mathbb Z[H]$-<module>, and those same generators generate it as a $\mathbb Z[K]$-ideal. Thus either chain condition passes through a finite factor.
If $K/H$ is infinite cyclic, choose $t\in K$ lifting its generator. Every element of $K$ has a unique form $ht^n$, and conjugation by $t$ gives an <automorphism> $\sigma$ of $\mathbb Z[H]$. Hence
$$
\mathbb Z[K]\cong\mathbb Z[H][t,t^{-1};\sigma],\qquad ta=\sigma(a)t.
$$
For the polynomial subring $C=\mathbb Z[H][t;\sigma]$, the condition $A+tA=A+At$ follows from invertibility of $\sigma$, so the proved <noncommutative Hilbert basis theorem> makes $C$ <Noetherian> on both sides. A <right ideal> $J$ of the <skew Laurent polynomial ring> is generated by $J\cap C$: for any $f\in J$, choose $m$ large enough that $ft^m\in C$, and recover $f=(ft^m)t^{-m}$. Finite generators of $J\cap C$ over $C$ therefore generate $J$ over the Laurent ring. Left multiplication by a large power of $t$ proves the left-handed statement. These are the Laurent-extension facts used in the induction.
Every factor of the <subnormal series> is covered by one of these two cases. Therefore
$$
\boxed{\mathbb Z[G]\text{ is left and right Noetherian}.}
$$
The argument proves the more general result that <group rings of poly-(cyclic or finite) groups are Noetherian> over any coefficient <ring> satisfying both chain conditions.
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