If $X$ is separated over $\operatorname{Spec} \Bbb Z$, is it separated over any other scheme? How is it connected that scheme $X$ is separated over $Y$ and over $\operatorname{Spec} \mathbb{Z}$, where $X \rightarrow Y$ is some scheme morphism?
 A: I guess you're asking if there is a connection between separatedness of a morphism $f:X\rightarrow Y$ and separatedness of $X$ (and\or $Y$) over $\mathrm{Spec}(\mathbf{Z})$. 
When one says that a scheme $X$ is separated, one means separated over $\mathrm{Spec}(\mathbf{Z})$, i.e., the unique morphism $X\rightarrow\mathrm{Spec}(\mathbf{Z})$ is separated. If $X$ is a separated scheme, then for any scheme $Y$, any morphism $f:X\rightarrow Y$ is separated. This is because $f$ can be factored as $\Gamma_f:X\rightarrow X\times_\mathbf{Z}Y$, the graph morphism, followed by the projection $X\times_\mathbf{Z}Y\rightarrow Y$. The first morphism is an immersion, hence separated, and the second morphism is a base change of the separated morphism $X\rightarrow\mathrm{Spec}(\mathbf{Z})$. So $f$ is a composite of separated morphisms, and therefore is itself separated. 
Thus a scheme $X$ is separated if and only if for all morphisms $f:X\rightarrow Y$, $f$ is separated. 
I'm not sure what else you're after. Taking $X=Y$ to be a scheme which is not separated (over $\mathrm{Spec}(\mathbf{Z})$, or equivalently, over some affine scheme, e.g. the spectrum of a field), then $\mathrm{id}_X:X\rightarrow X$ is trivially separated, but $X\rightarrow\mathrm{Spec}(\mathbf{Z})$ isn't. 
