[sp]I found solutions $(x,y) = (\pm4,3),\ (\pm4,5)$. I will look at mente oscura's solution to see if it can be adapted to show that these are the only solutions.[/sp]
Further thoughts:
[sp]The left side of the equation $(x^2-y^2)^2 = 16y+1$ is positive, so $y$ must be positive. Let $x-y = k$. Then the equation becomes $k^2(2y-k)^2 = 16y+1.$
Case 1: $k>0$. If two positive integers have a given sum (which in this case will be $2y$) then their product is minimised by taking one of them equal to $1$ and the other to be $2y-1$. So the minimum value of $k(2y-k)$ is $2y-1$, and therefore $k^2(2y-k)^2 \geqslant (2y-1)^2$. Thus $ (2y-1)^2 \leqslant 16y+1$. That simplifies to $y^2 \leqslant 5y$, so that $y \leqslant 5$.
Case 2: $k<0$. In this case, let $r=-k$, so that the equation becomes $r^2(2y+r)^2 = 16y+1$. This time, $r$ is a positive integer, so that $r(2y+r) > 2y$ and hence $r^2(2y+r)^2 > 4y^2$. Thus $4y^2 < 16y+1$ from which it follows that $y\leqslant4.$
This shows that there are no solutions with $y>5$.
BUT ...
in a PM, anemone pointed out that I overlooked the solutions $(x,y) = (\pm1,0)$. So altogether there are six solutions $(\pm1,0), (\pm4,3), (\pm4,5)$.[/sp]