This follows from Hilbert's basis theorem, which is valid for polynomial rings over any noetherian ring. But is there a more elementary proof, knowing that $\mathbb{Z}$ is a PID (even a Euclidean domain)?
[Math] $\mathbb{Z}[X]$ is noetherian
abstract-algebracommutative-algebra
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Take a Noetherian ring $R$. Then $R^n$ is a Noetherian module and all its submodules are Noetherian. For instance, PIDs are Noetherian, because all their ideals are principal, hence finitely generated. You could take $R = \mathbb Z$ or $R = k[x]$ (where $k$ is a field) and consider $M = R^n$. Looking at submodules of $M$, you get more Noetherian modules which are not rings.
Just to completely write out an example, let $R = k[x]$ and consider $\mathfrak m = \{ f(x) \in k[x] \, | \, f(0) = 0 \}$. Then $\mathfrak m \oplus R \trianglelefteq R^2$ is a Noetherian submodule of $R^2$. Its elements are pairs $(f(x), g(x))$ such that $f(0) = 0$.
(I guess this example is a ring without $1$, but for me rings always have a $1$ (because I don't like to think of the ideals of a unital ring as non-unital rings), so I thought it was a good enough example...)
Note that because of the classification of finitely generated modules over a PID (or more generally a Dedekind domain), all finitely generated $R$-modules over such rings will be isomorphic to $$ M \simeq R^{n-1} \oplus \mathfrak a \oplus \mathrm{Tor}(M) $$ where $\mathfrak a$ is a fractional ideal of $R$ (in the PID case, $\mathfrak a = R$) and $$ \mathrm{Tor}(M) \simeq R/\mathfrak p_1^{a_i} \oplus \cdots \oplus R/\mathfrak p_m^{a_m}, $$ so if you want something that is not a commutative ring with $1$, you need to take a ring which is not a Dedekind domain, in particular not a PID. Or if you stick with these rings, you need a non-finitely generated module.
Hope that helps,
Here is an outline of a proof, based on the last exercise in Atiyah-Macdonald:
Take a prime ideal $\mathfrak{p}\subset A$ of maximal height $m$. It is sufficient to show that $\mathfrak{p}[X]$ has height $m$ in $A[X]$, because $A[X]/\mathfrak{p}[X] \equiv (A/\mathfrak{p})[X]$ has dimension $1$.
The first step is to find an ideal $\mathfrak{a}=(a_1,\ldots , a_m) \subset \mathfrak{p}$ such that $\mathfrak{p}$ is minimal over $\mathfrak{a}$.* The next step is to show that $\mathfrak{p}[X]$ is minimal over $\mathfrak{a}[X]$.** This tells us that $\mathfrak{p}$ has height at most $m$,*** and it is easy to show that it has height at least $m$.
* and *** seem to require something like the Hauptidealsatz, while ** may require primary decomposition.
Note that this proof is quite straightforward when $A$ is a PID.
There are known counterexamples where the dimension of $A[X]$ can be as large as $2m+1$. See my example here.
Best Answer
Using the information you have about the ideals of $\mathbb Z$ and of polynomial rings in one variable over fields (both of which are PIDs), you can —with some work— describe all ideals of the ring, and then check the ACC by hand.