Question: Find all the positive integers $a$ such that $x^2+ax-1 = y^2$ has a solution in positive integers $(x,y)$.
Comments: It's easy to see that this equation rarely has a solution (in the sense that for a fixed $a$, $x^2+ax-1$ is a perfect square only for finitely many values of $x$). In fact, if $a$ is even then $x^2 \le x^2+ax-1 < (x+a/2)^2$, so $x^2+ax-1$ is almost never a perfect square. The problem is that I can't control the interval $[x^2,(x+a/2)^2)$ when $a$ grows. There's a similar argument for $a$ odd.
However, it is possible to find some families of such $a$'s. For instance, if $a$ is a perfect square, say $a=k^2$, then there exists the solution $(x,y) = (1,k)$.
In addition, if $x > 1$ then its prime power factors are $2$ and/or $p^\alpha$, where $p \equiv 1 \pmod 4$. In fact, if $p|x$ then the constraint of the equation modulo $p$ yields that $-1$ is a square.