I think I have an argument that might work. The goal is to prove that this is impossible. There are some gaps in it.

Let $X$ be a connected smooth proper scheme over $\mathbb Z$. Clearly $\Gamma(X,\mathcal O_X)=\mathbb Z$. (If the ring had zero-divsors, it would indicate $X$ reducible, impossible, or $X$ non-reduced, thus ramified, impossible. If it were a ring of integers of a number field it would give ramification at some prime.) Since $H^1(X,\mathcal O_X)$ is the tangent space of the Picard scheme, and the Picard scheme is trivial, $H^1(X,\mathcal O_X)$ is trivial. (This probably requires smoothness of the Picard scheme. I'm not sure if that holds.) I need to assume that $H^2(X,\mathcal O_X)$ is torsion-free. (I would think that smoothness over a scheme should imply locally free higher pushforwards, which over an affine scheme implies torsion-free cohomology, but I don't know. This is true in characteristic 0 by Deligne, but we are obviously not in characteristic 0 here.)

We have the exact sequence $0\to \mathcal O_X \to \mathcal O_X \to \mathcal O_X/p\to 0$, with the first map multiplication by $p$. Taking cohomology and filling in what we know, we get

$ 0 \to \mathbb Z \to \mathbb Z \to H^0(X, \mathcal O_X/p) \to 0 \to 0 \to H^1(X,\mathcal O_X/p) \to H^2(X,\mathcal O_X)\to H^2(X,\mathcal O_X) $

which since those are also the cohomology groups of $X_P$, gives $\Gamma(X_p,\mathcal O_{X_p})=\mathbb F_p$, $H^1(X_p,\mathcal O_{X_p})=0$.

Now let $Y\to X$ be a cyclic etale cover of degree $p$. Artin-Schreier on $X$ gives $H^1_{et}(X_P,\mathbb Z/p)=\mathbb Z/p$. Thus there is a unique connected etale degree-$p$ cover of $X_p$, so it's the one you get by tensoring over $\mathbb F_p$ with $\mathbb F_{p^p}$. Since $\Gamma(Y_p,\mathcal O_{Y_p})=\mathbb F_p$, it is connected, and is not the result of tensoring anything with $\mathbb F_{p^p}$. This is a contradiction.

No cyclic etale covers of degree $p$ $\implies$ no cyclic etale covers $\implies$ no etale covers. (since ever group has a cyclic subgroup.)