Existence of a finite extension of ℤ providing a finite extension of the primes Let $R$ be a ring (possibly noncommutative with zero-divisors). A non-unit and non-zero-divisor element $r \in R$ will be called irreducible if for all $a,b \in R$ such that $r=ab$, then $a$ or $b$ is a unit.
Many extensions $R$ of $\mathbb{Z}$ do not keep its set of irreducible elements, for example $2 = (1+i)(1-i)$ in $\mathbb{Z}[i]$. More generally (Chebotarev's density theorem) if $R$ is the ring of integers of a Galois extension of $\mathbb{Q}$ of degree $n$ then the prime numbers that completely split in $R$ have density $1/n$ (so that there are infinitely many ones).
Here are examples of extensions of $\mathbb{Z}$ keeping its irreducible elements:

*

*$\mathbb{Z}[X]$: every prime number is an irreducible polynomial, but $\mathbb{Z}[X]$ contains also infinitely many other irreducible polynomials (up to units),


*$\mathbb{Z} + \mathbb{Q}\epsilon$ (with $\epsilon^2=0$): it keeps the irreducibles of $\mathbb{Z}$ but there is no new ones up to units (see this comment).
Observe that the extensions of $\mathbb{Z}$ mentioned above (keeping its irreducible elements) either have infinitely many new irreducible elements (up to units) or have none (up to units).
In this answer Keith Conrad provides an example with finitely many new irreducible elements (up to unit):

Extend $\mathbf Z[x]$ by inverting all nonconstant irreducible
elements except for $x$. That gives you a UFD whose primes elements
are the ordinary prime numbers and $x$, up to units.

A finite extension of $\mathbb{Z}$ is a ring which is an extension of $\mathbb{Z}$ and a $\mathbb{Z}$-module of finite type. The example above is not.
Question: Is there a finite extension $R$ of $\mathbb{Z}$ keeping its irreducible elements, and also with new irreducible elements (up to units), but only finitely many (up to units)?
As Keith Conrad pointed out in this comment, if $R$ is an integral domain, then we are in the situation of Chebotarev's density theorem mentioned above. So $R$ must be noncommutative and/or with zero-divisors.

"up to units" means that if $r$ is an irreducible element and $u,v$ are units, then $r$ and $urv$ count for one.
 A: 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.
