Sharp bounds on the number of solutions of $X^{2}-\left( a^{2}+b^{2} \right) Y^{4}=-b^{2}$

We generalise and improve a result of Stoll, Walsh and Yuan by showing that there are at most two solutions in coprime positive integers of the equation in the title when $b=p^{m}$ where $m$ is a non-negative integer, $p$ is prime, $(a,p)=1$, $a^{2}+p^{2m}$ not a perfect square and $x^{2}- \left( a^{2}+p^{2m} \right) y^{2}=-1$ has an integer solution. This result is best possible. We also obtain best possible results for all positive integer solutions when $m=1$ and $2$. When $b$ is an arbitrary square with $(a,b)=1$ and $a^{2}+b^{2}$ not a perfect square, we are able to prove there are at most three solutions in coprime positive integers provided $x^{2}- \left( a^{2}+b^{2} \right) y^{2}=-1$ has an integer solution and $x^{2}- \left( a^{2}+b^{2} \right) y^{2}=-b^{2}$ has only one family of solutions. Our proof is based on a novel use of the hypergeometric method that may also be useful for other problems.
first draft for CNTA 2018. Comments welcome!