The Quillen-Suslin theorem states that any finitely generated projective module over $\mathbb{k}[x_1,...,x_m]$ is free, for $\mathbb{k}$ a field.
Is it known whether this statement is true in the case that $\mathbb{k}=\mathbb{Z}$, rather than a field?
Alternatively, a counter-example would be great.
On
As far as I know, this is true for PIDs and even without Noetherianity: it suffices for your base ring $A$ to be a Bézout domain. If you want a constructive proof, I think you can get away with assuming $A$ is a valuation domain or a finite-dimensional Bézout domain.
Recall that for a morphism of rings $A \rightarrow B$, a $B$-module $M$ is said to be extended from $A$ if there exists an $A$-module $M'$ such that $M \cong M' \otimes_A B$. Note that a free $B$-module is always extended from a (free) $A$-module. The big non-trivial result we need is the following:
Lequain-Simis: If $A$ is an arithmetic ring (i.e. $A_\mathfrak{p}$ has totally ordered ideals for each prime $\mathfrak{p}$), then every f.g. projective module of $A[x_1, \ldots, x_n]$ is extended from $A$.
This is proved, for example, in XVI.6 of H. Lombardi's Commutative Algebra: Constructive Methods: Finite Projective Modules, available here. Note that the abstract induction given there in 6.12 is adapted to a concrete induction 6.13 due to I. Yengui with the additional assumption that Krull dimension is finite.
When $B = A[X] := A[x_1, \ldots, x_n]$ we can say more about extended modules. The homomorphism $A[X] \rightarrow A$ sending polynomials to their constants gives us a way to send $A[X]$-modules to $A$-modules by extension of scalars. Extension of scalars preserves f.g. and projective objects. If $M$ is extended from $A$, then $M \cong M' \otimes_A A[X]$ so $M \otimes_{A[x]} A \cong M' \otimes_A A[X] \otimes_{A[x]} A \cong M' \otimes_A A \cong M'$. Thus if $M$ is extended from $A$, then extension of scalars of $M$ along $A[X] \rightarrow A$ is inverse to extension of scalars of $M$ along $A \rightarrow A[X]$, and in particular, if $M$ is f.g. projective and extended from $A$, then $M$ is extended from an f.g. projective $A$-module.
Our elementary remarks up to this point allow us to deduce the following:
Let $A$ be a ring which has f.g. projective modules free. Then $A[X]$ has f.g. projective modules free iff $A[X]$ has f.g. projective modules extended from $A$.
Applying this to the result of Lequain-Simis we thus see that any arithmetic ring having f.g. projectives free is a good Quillen-Suslin ring.
Two important classes of such rings come to mind.
(1) Chain rings, i.e. rings in which ideals are totally ordered. They have f.g. projectives free already because they are local, and local rings have all projectives free.
(2) Bézout domains (domains in which f.g. ideals are principal). To see that Bézout domains have f.g. projectives free, we´ll sketch a quick proof that Bézout domains are characterized as having f.g. submodules of free modules free. Indeed, if $A$ is Bézout and $M \subseteq F$ with $F$ free, then we can first assume $M \subseteq A^n$ by the f.g. hypothesis. The ideal $I$ generated by the first coordinate of $M$ is f.g. since $M$ is, and hence is principal by the Bézout assumption. Let $K = \ker(M \rightarrow I)$. Principal ideals of domains are free, so the exact sequence $0 \rightarrow K \rightarrow M \rightarrow I \rightarrow 0$ splits. Since $K$ is a f.g. submodule of $A^{n-1}$, we conclude by induction on $n$.
Finitely generated projective modules are free over $R[x_1,\dots,x_m]$ for any PID $R$. This was proved by Quillen in his original proof; I'm not sure about Suslin's proof. See Lam's Springer monograph "Serre's Problem on Projective Modules". (In fact all projective modules are free by a 1963 result of Bass.)