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?

Not an answer, just thinking out loud about the question of whether we can ever take $x,y$ relatively prime. 
We solve $x^2-y^2=a^3$ by writing $a^3=a_1a_2$ with $a_1,a_2$ of the same parity, then $x-y=a_1$, $x+y=a_2$, $2x=a_2+a_1$, $2y=a_2-a_1$, and we see any common divisor of $a_1,a_2$ other than perhaps $2$ is a common divisor of $x,y$. So we have $a_1=c^3$, $a_2=d^3$ with $c,d$ relatively prime, or else $a_1=2c^3$, $a_2=4d^3$, or else $a_1=4c^3$, $a_2=2d^3$. I'll just look at the first case. 
We have $2x=d^3+c^3$, $2y=d^3-c^3$, $c,d$ relatively prime. Then $$8b^2=(2x)^3-(2y)^3=(d^3+c^3)^3-(d^3-c^3)^3=6d^6c^3+2c^9$$ so $(2b)^2=c^3(3d^6+c^6)$. Since $c,d$ are coprime, the only possibilities for $\gcd(c^3,3d^6+c^6)$ are $1$ and $3$. I'll just look at the first case. 
Now $c^3$ and $3d^6+c^6$ must be squares, so $c=e^2$ and $3d^6+e^{12}=f^2$. So $$(f-e^6)(f+e^6)=3d^6$$ If $f-e^6,f+e^6$ are coprime (and they must be pretty nearly coprime, if you trace back through to the coprimality of $c,d$), then $f-e^6=g^6$, $f+e^6=3h^6$, or else $f-e^3=3g^6$, $f+e^6=h^6$. Taking, as usual, just the first case, we get $$g^6+2e^6=3h^6$$ 
That's as far as I go. There is the trivial solution, $e=g=h=1$. That rules out using congruences to show there aren't any solutions, but I suspect there aren't any other coprime solutions. Maybe someone can actually prove this, and clean up the cases I haven't discussed. 
