Families of curves over number fields

Is there known a number field $K$ and a curve $F(x, y) \in K(t)[x, y]$ such that $F(x, y)$ does not have points over the field of rational functions $K(t)$ but for all but finitely many positive integer values of $t$ the respective specialization has a point over $K$?

(I expect the answer to be "such a $K$ is not known" and the problem of similar difficulty as that for $K = \mathbb{Q}$ but have not seen an expert comment on it so far.)

Thank you.

Assuming the parity conjecture (known to be a consequence of the BSD conjecture) for "congruent number curves" $E_d: d \cdot Y^2 = X^3 - X$, you can even take $K = {\bf Q}$ and $$ F(x,y) = 4x^2 + 1 + (8t-1) x y^2. $$

The curve $F(x,y) = 0$ is birational with the elliptic curve $(8t-1) Y^2 = X^3 + 4X$ (let $x=1/X$ and $y=Y/X$), which in turn is 2-isogenous with $E_{8t-1}: (8t-1) Y^2 = X^3 - X$. The only ${\bf Q}(t)$-rational points of the curve $(8t-1) Y^2 = X^3 + 4X$ are the torsion points at $(0,0)$ and infinity, and I chose coordinates $X,Y$ that put both of these points on the line at infinity, so there are no finite solutions of $F(x,y) = 0$. On the other hand, if $t$ is a positive integer then $|8t-1| \equiv 7 \bmod 8$, so $E_{8t-1}$ has sign $-1$, whence under BSD it has odd (and thus positive) rank over $\bf Q$, so infinitely many rational points.

[This example has no exceptional $t$, but it's clear a priori that if there's an example with only finitely many exceptional $t$ then one can translate $t$ to get a "new" $F$ that has none.]