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.]