As promised, here is my solution based on the Davis trick. First, there is a very general construction of PL aspherical 4-manifolds (it also works in higher dimensions). Start with a finite aspherical 2-dimensional CW complex $W$. Up to homotopy, $W$ always embeds in the Euclidean 4-space $E^4$ (I think, this is due to Stallings). Take such an embedding and let $N=N(W)$ denote a regular neighborhood of $W$ in $E^4$. Now, apply "Davis trick" to $N$: Introduce a reflection orbifold structure on the boundary of $N$ such that the corresponding stratification of the boundary is dual to a triangulation of $\partial N$. The resulting orbifold ${\mathcal O}$ is very good (admits a finite orientable manifold-covering $M\to {\mathcal O}$) and its universal covering (same for $M$ and for ${\mathcal O}$) is contractible. As a bonus, $\pi_1(W)$ embeds in $\pi_1(M)< \pi_1({\mathcal O})$. For details see

Mess, Geoffrey, Examples of Poincaré duality groups, Proc. Am. Math. Soc. 110, No. 4, 1145-1146 (1990). ZBL0709.57025.

and, of course, the original paper by Mike Davis from 1983. (Actually, it was Bill Thurston who came up with this trick in the context of 3-manifolds: He used it for the proof of his hyperbolization theorem.) This construction allows one to embed 2-dimensional finitely presented groups with "exotic properties" in fundamental groups of closed aspherical PL manifolds.

I will use a relative version of this construction. Start with a closed connected oriented surface of genus $\ge 1$; I'll take the torus $T^2$. Let $E\to T^2$ be the 2-disk bundle over $T^2$ with the Euler number $\pm 1$. The boundary of the 4-manifold $E$ is a 3-dimensional Nil-manifold: The total space of a nontrivial circle bundle over the torus. The group $\pi_1(\partial E)$ has two generators $a, b$, and $\pi_1(\partial E)$ has the presentation $$ \langle a, b| [a,b]=t, [a,t]=1, [b,t]=1\rangle. $$ Represent $a, b$ by simple disjoint loops $\alpha, \beta$ in $\partial E$. Now, take your favorite finite 2-dimensional aspherical complex $W$ whose fundamental group is nontrivial and has no proper finite index subgroups (I care only about the homotopy type of $W$). The standard example is the presentation complex of Higman group. But there are many other examples. As before, embed $W$ in $E^4$, take a regular neighborhood $N$ of $W$ in $E^4$. Then $\pi_1(\partial N)$ maps nontrivially to $\pi_1(W)$. Pick two simple loops $\alpha', \beta'\subset \partial N$ which map nontrivially to $\pi_1(W)$ (you can take the same loop).

Now, take two copies $N_a, N_b$ of $N$ and attach them to $E$ by identifying a regular neighborhood of $\alpha'$ to that of $\alpha$ for $N_a$ and identifying a regular neighborhood of $\beta'$ to that of $\beta$ for $N_b$. The result is a compact PL aspherical 4-manifold with boundary $Z$. The $\pi_1(Z)$ is an amalgam of $\pi_1(E)\cong {\mathbb Z}^2$ with two copies of $\pi_1(N)$ (along infinite cyclic subgroups). For each homomorphism to a finite group $$ \phi: \pi_1(Z)\to \Phi $$ the subgroups $\pi_1(N_a), \pi_1(N_b)$ will have to map trivially. Hence, $a$ and $b$ will have to map trivially as well. Since $a, b$ generate $\pi_1(E)$, $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Now, apply Davis trick to $Z$. The result is an orbifold ${\mathcal O}$. Since $Z$ was aspherical, so is ${\mathcal O}$ (i.e. it has contractible universal covering space).

Take a finite orientable manifold-covering $M\to {\mathcal O}$. Then $M$, of course, has contractible (hence, spin) universal covering. I claim that $M$ has no finite spin-covering spaces. Indeed, for each finite-sheeted covering $p: M'\to M$, the manifold $int(Z)\subset M$ has to lift trivially; more precisely, $p$ restricts to a trivial covering $$ p^{-1}(int Z)\to int Z.$$ This is because $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Thus, $M'$ contains a copy of $E$. In particular, $M'$ contains a 2-torus with odd self-intersection, i.e. the intersection form of $M'$ is not even, i.e. $M'$ is not spin.

I was working in the PL category but in dimension 4, PL is the same DIFF, so you get a smooth example as well.