$\DeclareMathOperator\Spec{Spec}$Does there exist a scheme $X/{\operatorname{Spec}(\mathbb{Z})}$ such that $\pi_1^\text{ét}(X)=\smash{\hat{\mathbb{Z}}}^2$?

It's well known that $\pi_1^\text{ét}(\Spec(\mathbb{F}_p))=\hat{\mathbb{Z}}$, for $p$ a prime. However, the scheme $$X_{p,q}:=\Spec(\mathbb{F}_p)\times_{\Spec(\mathbb{Z})}\Spec(\mathbb{F}_q)$$ is empty if $p\neq q$, and isomorphic to $\Spec(\mathbb{F}_p)$ if $p=q$, so neither work. The idea was to use the natural map, $$\pi_1^\text{ét}(X\times_{S}Y)\rightarrow \pi_1^\text{ét}(X)\times_{\pi_1^\text{ét}(S)} \pi_1^\text{ét}(Y)$$ somehow. So does such a scheme exist? 

Edit:

As @Will Chen pointed out in the comments. $X= $an elliptic curve or $\mathbb{G}_m^2$ work. However, that's not really the question I meant to ask. I'd like the map to be an embedding$$X\hookrightarrow\text{Spec}(\mathbb{Z})$$ The reason is Spec$(\mathbb{Z})$ has cohomological dimension $3$ (up to $2-$torsion). And $$\text{Spec}(\mathbb{F}_p)\hookrightarrow\text{Spec}(\mathbb{Z})$$ is an embedding, and Spec$(\mathbb{F}_p)$ has cohomological dimension 1. So also, $X$ with cd$(X)=2$.