Rank of jacobians of twists of hyperelliptic curves of genus one

If you are willing to assume finiteness of Sha, then the answer is "yes".

The first claim, that for infinitely many $d$ (square-free, presumably?) this has a rational point, is easy and does not need any conjectural input: start with any $u$, $v$ such that $F_{a,b}(u_0,v_0)$ is non-zero. Then by writing $F_{a,b}(u_0,v_0)$ as a square times the square-free part, you immediately see that there exists a unique square-free $d\in \mathbb{Z}$ such that $F_{a,b}(u_0,v_0)/d$ is a square, say $z_0^2$, so that $(u_0,v_0,z_0)$ is a rational point on $dz^2=F_{a,b}(u,v)$.

For the second part, observe, e.g. by using the explicit equation for the Jacobian that you have given, that the Jacobian of the twist by $d$ is the twist by $d$ of the Jacobian. Since these Jacobians have full $2$-torsion, it follows from work of Kane that infinitely many of these have $2$-Selmer of dimension $3$, and if Sha is finite, then $3$-dimensional $2$-Selmer implies rank $1$.