Hey Bjorn. Let me try for a counterexample. Consider a hypersurface in projective $N$-space, defined by one degree 2 equation with integral coefficients. When is such a gadget smooth? Well the partial derivatives are all linear and we have $N+1$ of them, so we want some $(N+1)$ times $(N+1)$ matrix to have non-zero determinant mod $p$ for all $p$, so we want the determinant to be +-1. The determinant we're taking is that of a symmetric matrix with even entries down the diagonal (because the derivative of $X^2$ is $2X$) and conversely every symmetric integer matrix with even entries down the diagonal comes from a projective quadric hypersurface. So aren't we now looking for a positive-definite (to stop there being any Q-points or R-points) even unimodular lattice?
So in conclusion I think that the hypersurface cut out by the quadratic form associated in this way to e.g. the $E_8$ lattice or the Leech lattice gives a counterexample!
That is way too ambitious, I think, considering what is known about classification of algebraic varieties. Ignoring the trivial, for these purposes, case of curves, the next and best studied case is algebraic surfaces. For them, the classification is classical and has been known for almost a century. So that is a good first case to consider.
So you are asking: what are the smooth projective surfaces $X$ with $\pi_1(X)=1$, and in particular with regularity $q=h^1(\mathcal O_X)=0$. Well, pick up your favorite text, Shafarevich etc. or Barth-(Hulek-)-Peters-van de Ven, or whatever, and go through the list.
Kodaira dimension $-\infty$: here you get all rational surfaces, and only these.
Kodaira dimension $0$: K3 surfaces.
Kodaira dimension $1$, elliptic surfaces $X\to C$ (a general fiber is elliptic). Clearly, $C$ must be $\mathbb P^1$. But actually getting $\pi_1(X)=1$ seems like not a completely trivial condition, something to think about.
Kodaira dimension $2$, i.e. surfaces of general type. Well, the examples of simply connected surfaces of general type are highly prized, especially if they have $p_g=h^0(\omega_X)=0$. Many such surfaces are known (e.g. http://en.wikipedia.org/wiki/Barlow_surface, some Godeaux surfaces, some Campedelli surfaces) but a complete classification? Not even close. Like I said, this is too ambitious for the present state of knowledge.
Best Answer
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.)