Diophantine equations with cubic roots. 
Find all positive, integral $x, y$ such that 
  $$x^3-y^3=xy+61$$

Little work:
$$(x-y)(x^2+xy+y^2)-xy=61$$
Also $x^3-xy-y^3$ seems to to be unable to be "completed by the square," or factored, so I'm stuck. Thanks!

More generally,
suppose
$x^3-y^3=xy+m
$
where $x, y, m > 0$.
I will show that
$m \ge x^2+y^2$,
$x \ge y+1$,
$x \lt \sqrt{m}$
and
$y
\le \frac12(\sqrt{2m-1}-1)
$.
If $m=61$ then
$x \le 7$
and
$y 
\le \frac12(\sqrt{120}-1)
\approx 4.97
$
so
$y \le 4
$
and
$x \le \sqrt{61-y^2}
$.
We must have
$x > y$
so that
$x \ge y+1$
and
$xy+m
=(x-y)(x^2+xy+y^2)
\ge x^2+xy+y^2
$
so
$m
\ge x^2+y^2
$.
Therefore
$m \gt x^2
$
and
$x \ge y+1$
so that
$m
\ge (y+1)^2+y^2
=2y^2+2y+1
$
or
$2m-1
\ge 4y^2+4y+1
=(2y+1)^2
$
so
$y
\le \frac12(\sqrt{2m-1}-1)
$.
