maybe its work ! for $gcd(xy,10)=1$ obviously we have contradiction .if $5|x$ we have : $50x'^3-y^3=2x'-y$ it means $x-y|49x^2-1$ also should $y^2+x^2+xy-1|49x^2-1$ which you can say: $(y^2+x^2+xy-1)k=49x^2-1$ which Quadratic Equation and easy for solving. for $2|y$ also approximately similar to this.

Bump.... One can prove that $$y(x+y)(x-y)|42(x^2-1)$$ By taking mod $y$,$x+y$,$x-y$.

Rewrite the equation as $2x^3-5y^3=2x-5y $. Consider modulo $2x-5y $ because we have symmetry. So we have $$2x^3\equiv 5y^3\pmod {2x-5y} $$$$25\cdot 2x^3\equiv (5y)^3\equiv 8x^3\pmod {2x-5y} $$ Indeed we get $2x-5y\mid 21$. Now we need to consider this cases. Let $P(k) $ denote the diophantine $2x^3-5y^3=2x-5y=k $ so we need to consider the cases $P (\pm 1), P (\pm 3), P (\pm 7),P (\pm 21) $. Notice we can rewrite $P (k) $ as $P (k):\frac {(5y+k)^3}{4}+5y^3=k $ or alternatively $P (k):y (15k^2+75ky+145y^2)=4k-k^3$. Also notice that if $2x=5y $ this can be checked since implies $y\mid 2$ and $x\mid 5$. $P (\pm 1)\implies 5y\mid 3$ contradiction $P (\pm 3)\implies y\mid 3$ $P (\pm 7)\implies y\mid 63$ $P (\pm 21)\implies 5y\mid 9177$ contradiction. So this give us the solutions $(x,y)=(0,\pm 1),(\pm 1,0),(\pm 4,\pm 3),(\pm 1,\mp 1),(0,0) $.