Solve in integers the equation $\sqrt{x^3-3xy^2+2y^3}=\sqrt[3]{13x+8}$ 
Solve in integers the equation 
  $$\sqrt{x^3-3xy^2+2y^3}=\sqrt[3]{13x+8}$$

My work so far:
I used www.wolframalpha.com. Then $x=9,y=8 -$ solution.
My attempt:
1) Let $\sqrt{x^3-3xy^2+2y^3}=a, \sqrt[3]{13x+8}=b$. Then 
$$\begin{cases}
a=b\\
a^2-b^3=x^3-3xy^2+2y^3-13x-8 \in \mathbb Z
\end{cases}$$
2) $$(x^3-3xy^2+2y^3)^3=(13x+8)^2$$
Addition:
We have $$A^6\cdot B^6=(13x+8)^2$$
and $$n^6 \equiv 0, \pm 1 (\bmod 13)$$
 A: There are no integer solutions with $x<0$, since $13x+8<0$ for such $x$. When $0\leq x\leq 8$ then $13x+8$ is not an integer cubic. Since the latter is easily seen to be a necessary condition it follows that there are no solutions with $x$ in this range. On the other hand you already have found the solution $(9,8)$. In the following I shall prove that there are no solutions with $x\geq10$. 
I'm considering $x\geq10$ as "primary variable" and write $y:=x-t$ with a new  integer variable $t$. The equation at stake then reads
$$a:=\sqrt{t^2(3x-2t)}=\root 3\of{13x+8}=:b\ .$$
If $t\leq-1$ then
$$a=|t|\sqrt{3x+2|t|}\geq\sqrt{3x}\ .$$
For $x=10$ we therefore have $a^3\geq\bigl(\sqrt{3x}\bigr)^3=30^{3/2}>164>138=13x+8=b^3$, and things get worse for larger $x$.
If $t=1$ and $x=10$ then $a=\sqrt{3x-2}=\sqrt{28}$.
For $x=10$ we therefore have $a^3\geq28^{3/2}>148>138=b^3$, and things get worse for larger $x$.
If $2\leq t\leq x$ then
$$a\geq2\sqrt{3x-2t}\geq2\sqrt{x}\ .$$
For $x=10$ we therefore have $a^3\geq8\cdot10^{3/2}>252>138=b^3$, and things get worse for larger $x$.
Finally, if $2x\leq 2t\leq 3x-1$ then
$$a=\sqrt{t^2(3x-2t)}=t\sqrt{3x-2t}\geq x\ .$$
For $x=10$ we therefore have $a^3\geq10^3>138$, and things get worse for larger $x$.
