My question is the following:$\newcommand{\Q}{\Bbb Q} \newcommand{\Z}{\Bbb Z}$

What is known about number fields $K$ fulfilling the condition $C_{g,K}$ "there is a smooth projective curve of genus $g$ over $K$, having everywhere good reduction" for some $g \geq 1$ ?

By the work of Fontaine and Abrashkin, it is known that for every $g>0$, the field $K = \Q$ does not satisfy $C_{g,\Q}$. Notice that the condition $C_{g,K}$ (for some $g>0$) implies, by the functorial properties of the Jacobian, the property $A_K$ "there is a non-zero abelian variety over $K$ having good reduction everywhere". (Using the theory of Néron models, it should be equivalent to state that there is no non-trivial abelian scheme over the ring of integers $O_K$ (see here, where it is also explained that a CM abelian variety has potentially "good reduction everywhere")).

Thus a closely related question is:

Do we expect the existence of (quadratic?) number fields $K \neq \Q$ such that the assertion $A_K$ does not hold (so that in particular, there is no smooth projective curve of genus $>0$ over $K$ with everywhere good reduction)?

For instance, what happens if $K = \Bbb Q\left(\sqrt{-2}\right)$ or $K = \Bbb Q(\sqrt{2})$?

It is mentioned here that there is no elliptic curve over $\Q(\sqrt 2)$ with everywhere good reduction. The same happens with $\Q(i)$, see this answer. Examples of abelian surfaces with everywhere good reduction (i.e. the opposite of what I'm looking for) are mentioned here.

By some analogy discussed here, it may be useful to note that $\Q(i)$ and $\Q(\sqrt 2)$ have no non-trivial unramified extension (see here).

It was asked here whether *every* number field $K \neq \Q$ satisfies $C_{g,K}$ for some $g>0$, but the answer is only very partial.
Furtherfore, it is explained here that for every $g \geq 0$, there is *some* number field $K$ such that $C_{g,K}$ holds — this is different from my question, where I want $K$ to be fixed at the beginning.