This post started as some minor observations, but I believe it now contains a full proof that there is no such ring. Throughout, we let $R$ be an example of the kind wanted.

**Observation 1**: $R$ is indecomposable.

*Proof of observation 1*: Write $R=S_1\times S_2$, with $S_1,S_2\neq 0$. Suppose, by way of contradiction, that $S_1$ has finite characteristic $m>1$. Let $p$ be any integer prime divisor of $m$. Then $p=(p,p)$ is not irreducible (since $(1,0)(p,p)=0$), contradicting our assumption that the primes of $\mathbb{Z}$ stay irreducible in $R$ (and using your definition of irreducible, which includes not being a zero-divisor).

Thus, each $S_i$ has characteristic $0$. Moreover, the argument above shows that each integer prime cannot be a zero-divisor in any $S_i$.

If an integer prime $p$ is a unit in $S_i$, then $S_i$ contains a copy of $\mathbb{Z}[1/p]$, which is not finitely generated as a $\mathbb{Z}$-module, contradicting the fact that submodules of finitely generated $\mathbb{Z}$-modules are finitely generated.

Also, if $p$ is not irreducible in $S_1$ (or, by symmetry $S_2$), say $p=ab$ where $a,b$ are nonunits, then $(p,p)=(a,p)(b,1)$ is a product into nonunits, contradicting the irreducibility of the primes of $\mathbb{Z}$.

The elements $\{(p,1)\, :\, p>1\text{ prime}\}$ are not units, not zero-divisors, and are easily shown to be irreducible in $R$. They are not associate to each other, nor to the integer primes $(p,p)\in R$. Thus, this would contradict our assumption of having only finitely many new primes (up to associates).

**Observation 2**: If $r\in R$, then the subring $\mathbb{Z}[r]\subseteq R$ is isomorphic to $\mathbb{Z}[x]/I$ where $I$ is a principal ideal generated by a monic polynomial $f$.

*Proof of observation 2*: Since $\mathbb{Z}[r]$ is a $\mathbb{Z}$-submodule of $R$, it must be finitely generated. Hence, $r$ satisfies some monic polynomial. Let $f\in \mathbb{Z}[x]$ be the monic polynomial of smallest degree satisfied by $r$.

Now, $\mathbb{Z}[r]\cong \mathbb{Z}[x]/I$ where $I$ is the ideal of polynomials satisfied by $r$. It thus suffices to show that $I=f\mathbb{Z}[x]$. Supposing otherwise, let $g\in I-f\mathbb{Z}[x]$.

Working over $\mathbb{Q}[x]$ for a moment, take $d=\gcd(f,g)$, a monic polynomial. Since $f$ is monic, we know that $d\in \mathbb{Z}[x]$. By the extended Euclidean algorithm, we can write $d$ as a $\mathbb{Q}[x]$-linear combination of $f$ and $g$. Hence $cd\in I$ for some minimal $c\in \mathbb{Z}_{>0}$. We know $c\neq 1$ by the minimality of the degree of $f$ (since $d$ is monic).

Let $p$ be any prime dividing $c$. Then $p\cdot (c/p)d(r)=0$, and hence $p$ is a zero-divisor in $R$, which is a contradiction.

**Observation 3**: The monic polynomial $f$ is a power of an irreducible polynomial $q\in\mathbb{Z}[x]$.

*Proof of observation 3*: Assume $f$ is not a power of an irreducible, so $f=gh$ over $\mathbb{Z}[x]$, where $\gcd(g,h)=1$ and $\deg(g),\deg(h)\geq 1$.

By an argument used previously, we can find some constant $c\in \mathbb{Z}_{>0}$ that is a $\mathbb{Z}[x]$-linear combination of $g$ and $h$, say $c=gg'+hh'$.

Let $J=g\mathbb{Z}[x]$ and $K=h\mathbb{Z}[x]$. Let $S_1=\mathbb{Z}[x]/J$ and $S_2=\mathbb{Z}[x]/K$. Consider the map $\varphi\colon \mathbb{Z}[x]/I\to S_1\times S_2$ where $a+I\mapsto (a+J,a+K)$. This is an injective ring homomorphism. Thus, we can view $\mathbb{Z}[r]$ as a subring of this direct product. Moreover the image of $\mathbb{Z}[r]$ contains the elements $(c+J,0+K)$ and $(0+J,c+K)$. (Indeed, $h(r)h'(r)\mapsto (hh'+J,hh'+K) = (c+J,0+K)$.) Since the image of $\mathbb{Z}[r]$ contains $(1+J,1+K)$, we see (by a trivial use of Dirichlet's theorem) that the image of $\mathbb{Z}[r]$ contains elements of the form $(p+J,1+K)$ and $(1+J,p+K)$, with $p$ an integer prime. Thus (suppressing the coset notation) we have $(p,p)=(p,1)(1,p)$ in this image. Back inside $\mathbb{Z}[r]$, write this factorization as $p=ab$. Without loss of generality, $a$ is a unit in $R$, with inverse $a^{-1}$.

The ring $\mathbb{Z}[r,a^{-1}]$ is a finitely generated $\mathbb{Z}$-module (being a $\mathbb{Z}$-submodule of $R$), and it naturally maps onto $(\mathbb{Z}[x]/J)[1/p]$, which is not a finitely generated $\mathbb{Z}$-module, which is a contradiction.

**Observation 4**: $\deg(q)=1$.

*Proof of observation 4*: If $\deg(q)>1$, then there are infinitely many primes $p$ that factor nontrivially in $\mathbb{Z}[x]/(q)$ (by an argument supplied by John Voight) and such a factorization lifts to $\mathbb{Z}[x]/(q^n)\cong \mathbb{Z}[r]$; see this link for the full argument.

Now, if such a prime $p$ is to be irreducible in $R$, one of those factors must become a unit $u$ in $R$. But then $\mathbb{Z}[r,u^{-1}]$ is a commutative ring, mapping to $(\mathbb{Z}[x]/(q))[a^{-1}]$ (where $a$ is that corresponding nontrivial factor, but modulo $q$ rather than $q^n$), which is not a finitely generated $\mathbb{Z}$-module, giving us a contradiction, as before.

**Observation 5**: If $J$ is the set of nilpotent elements in $R$, then $J$ is a (nilpotent) ideal.

*Proof of observation 5*: An arbitrary element $r\in R$ satisfies $q^n$, with $q(x)=x-k\in \mathbb{Z}[x]$, and hence $r=k-t$ where $t$ is nilpotent of index $n$. Therefore, $r(k^{n-1}+k^{n-2}t + \cdots + t^{n-1})=k^n$. If $k\neq 0$, then $k^n$ is not a zero-divisor (since all the primes of $\mathbb{Z}$ are not zero divisors), and hence $r$ is not a zero-divisor. Thus, the zero-divisors are exactly the nilpotent elements.

Next, if $t\in R$ is nilpotent, say of index $n\geq 1$, and $r\in R$ is arbitrary then $t^{n-1}(tr)=0$ with $t^{n-1}\neq 0$. Hence $tr$ is a zero-divisor, hence nilpotent. Thus $J$ is closed by right multiplication from $R$; and by left multiplication by a symmetric argument.

In particular $J$ is closed under multiplication. By the paper Rings in which nilpotents form a subring by Janez Ster (arXiv version here), we know that $J$ is also closed under addition, since $R$ satisfies Koethe's conjecture. (Quick argument: The Jacobson radical $J(R)$ is nilpotent, since tensoring up to $\mathbb{Q}$ keeps it in a finite-dimensional $\mathbb{Q}$-algebra.) Thus, $J$ is an ideal.

**Observation 6**: The type of ring we want cannot exist.

*Proof of observation 6*: Fix a $\mathbb{Z}$-basis for $J$, say $t_1,\ldots, t_k$. Then a $\mathbb{Z}$-basis for $R$ is $1,t_1,\ldots, t_k$ (since every element of $R$ is an integer shift from a nilpotent, by observation 4).

The units of $R$ are exactly $\pm 1+t$ for some $t\in J$. Thus, the associates of a prime $p\in \mathbb{Z}$ are of the form $\pm p+t'$ where $t'\in J$ is divisible by $p$. Thus $p+t_1$ is not associate to any of the integer primes. A similar argument shows that $p+t_1$ and $p'+t_1$ are not associate, for distinct primes $p,p'$. Further, $p+t_1$ is not a unit, and not a nilpotent (hence not a zero-divisor). It thus suffices to show that $p+t_1$ is irreducible, and then we'll have infinitely many "new" nonassociate irreducibles.

If $p+t_1$ factors as $(a+t)(b+t')$ with $a,b\in \mathbb{Z}$ and $t,t'\in J$, then $ab=p$. Hence, without loss of generality, $a=\pm 1$. Thus, $a+t$ is a unit, so every factorization is trivial.

finiteextension of $\mathbb{Z}$. Now I wonder whether the last part of the question (i.e. "but only finitely many (up to units)") is automatic for a finite extension of $\mathbb{Z}$... $\endgroup$4more comments