Solving $x^3-y^3=xy+61$ in integers

Put $x+y=:u$, $x-y=:v$; then $u$ and $v$ are integers of the same parity. In the new variables the equation becomes $${1\over8}(6u^2v+2v^3)={1\over4}(u^2-v^2)+61\ ,$$ which can be written as $$27 u^2+(3v+2)^2={6584\over3 v-1}\ .$$ It follows that $v>1$ and that $3v-1$ must divide $6584=8\cdot 823$; whence $v\in\{1,3,549,2195\}$. $549$ and $2195$ are obviously too large, and $v:=3$ leeds to $u^2=26$ which has no integer solutions. It remains $v=1$ which leads to $u^2=121$ or $u=\pm11$. Therefore we only have the two solutions $(x_1,y_1):=(6,5)$ and $(x_2,y_2):=(-5,-6)$.

I believe the only answers in the integers are $(6,5)$ and $(-5,-6)$.

If $x$ and $y$ are integers, then we can let $y=x-k$, where $k$ is an integer. Then the equation can be written as a quadratic in $x$: $$k(x^2+x(x-k)+(x-k)^2)-x(x-k)-61 = 0$$ The solutions to this are: $$x = \frac{k(3k-1) \pm\sqrt{(1-3k)(k^3+k^2-244)}}{2(3k-1)}$$ The term under the square root is only positive for $k=1,2,3,4,5$, and checking the results for each of these shows that only $k=1$ results in integer solutions (listed above).

If we restrict ourselves to positive solutions, we can do it by brute force. In that case clearly $x>y>0$.

If $x-y=1$, we have $x^2+y^2=61$, which has the unique solution $x=6,y=5$.

If $x-y=2$, we have $2(x^2+xy+y^2)=xy+61$, or $2(x^2+y^2)+xy=61$; clearly $x$ and $y$ must both be odd, with $x^2+y^2<30$. The only candidate is $x=3,y=1$, which clearly doesn't work.

If $x-y=3$, we have $3(x^2+y^2)+2xy=61$; here $x$ and $y$ must have opposite parity, with $x^2+y^2<20$, so $x=4,y=1$ is the only candidate, and it doesn't (quite) work.

If $x-y=4$, we have $4(x^2+y^2)+3xy=61$, so $x$ and $y$ must be odd, and $x^2+y^2<15$; there are no solutions in this case.

Suppose that $x-y\ge 4$. Then $(x-y)(x^2+xy+y^2)\ge 4(5^2+5\cdot1+1^2)>61$, so there are no solutions.