Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e.,
Positive integers $(x,y)$ such that
$x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$?
Finding infinite number of pairs is no problem, as in:
$( 2^{6j+1} \cdot 3^{6k} \cdot 5 , 2^{6j+1} \cdot 3^{6k+1} )$ for any integral $j,k \geq 0$ 
But how would you determine the exhaustive list?

$ x^2 - y^2 = a^3 $ is fairly trivial, because clearly for every integer solution there are integers $ p, q, r, s $ with $p$ and $q$ squarefree and coprime such that $ x + y = p q^2 r^3 $ and $ x - y = p^2 q s^3 $, so that:
$ 2 x = p q (q r^3 + p s^3) $
$ 2 y = p q (q r^3 - p s^3) $
$ a = p q r s $
and these satisfy the equation identically, with $x, y$ integers iff either $p q$ is even or $p, q, r, s$ all odd.
(or $r, s$ both even, although for "reduced" solutions with no integer $t > 1$ such that t^3 divides $x, y$ and t^2 divides $a$ we can assume that $gcd(r, s) = 1$, which we do hereafter ..)
Plugging these relations into $(2 x)^3 - (2 y)^3 = 8 b^2 $ gives $ p^4 q^3 (3 q^2 r^6 s^3 + p^2 s^9) = 4 b^2 $.
Noting that $p$ and $q$ are coprime and squarefree we then conclude $2 b = p^2 q^2 s B$ for some integer $B$, so that $ s (3 q^2 r^6 + p^2 s^6) = q B^2 $.
From this we see that $q$ divides $p^2 s^7$ and hence, since $gcd(p, q) = 1$ and $q$ is squarefree, that $q$ divides $s$.
Thus with $s = q S$ and $B = q C$ we obtain $ S (3 r^6 + p^2 q^4 S^6) = C^2 $.
Then, taking $S = u v^2$ with $u$ squarefree, we conclude that $C = u v D$ for some integer $D$ so that finally $3 r^6 + (p q^2 u^3 v^6)^2 = u D^2 $
In this $u$ divides $3 r^6$; but since $gcd(r, s) = 1$ and $u$ divides $s$ we conclude that $u$ divides 3.
Thus, since $u$ > 0, we must have $u = 1$ or $u = 3$, which reduces the problem to one of the following respectively:
$3 r^6 + M^2 = N^2$   ($u = 1$)
$ r^6 + 3 M^2 = N^2$  ($u = 3$)
Now $X^2 + 3 Y^2 = Z^2$ has general integer solution $X, Y, Z = k(m^2 - 3 n^2), 2 k m n, k(m^2 + 3 n^2)$ with $gcd(m, n) = 1$ and $m + n$ odd.
So the cases require either $2 k m n = r^3$ or $k(m^2 - 3 n^2) = r^3$, each of which is trivial by suitable choice of $k$ (although arguably if you want explicit values of m, n the second is not so trivial).
