Try : Let's say $d=(x,y); x=du,y=dv,(u,v)=1$ then the equation gives $d^3(u^6-v^6)=2016uv^2$. Now one can see, for $k|2016$, $u^6-v^6=ku^av^b\ ;a\in\{0,1\},b\in\{0,1,2\}$ has no solution unless $a=b=0$.
Then we only need to consider cases for $k|2016;\ u^6-v^6=k;\ d^3=\frac{2016}{k}uv^2;\ u>v$. Is my argument correct?
On
$x^6-y^6=2016 x y^2$
Clearly x and y must have a common divisors and we can write:
$(x-y)(x+y)(x^2+xy+y^2)(x^2-xy+y^2)=2016 x y^2$
$2016$can be written as the product of different factors such as : $7 . 9. 32=8 . 7 . 36=2 . 4. 252= 4 . 8 . 7. 9=4 . 12 . 6 . 7= . . .$
Among these orders only $4 .12 . 6 . 7 y^2 x$ gives the solution:
$x-y=4$
$x+y=12$
Which gives $x=8$ and $y=4$
Sixth powers grow much faster than cubes, so for given $X$, there has to be a maximum $Y$. It also seems plausible that there might be a point where $X^6 –(X-1)^6 > 2016X(X-1)^2$, hence the search is finished.
A trivial program, thrown together just to gage the size of the calculation, shows that cut-off is $X=19$, after only 53 $(X,Y)$ pairs.
The only solution is $(X,Y) = (8,4)$