Find all natural numbers $x$ and $y$ such that $$ \frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x^2 - y^2}} \in \mathbb{N}. $$
My approach: $x>y$ because otherwise the radical would be negative and we can see that $(1;0)$ is a solution to the equation. How do we prove that there are other solutions?
Let $$ k=\frac{\sqrt{x} + \sqrt{y}}{\sqrt[3]{x^2 - y^2}} $$ for some positive integer $k$. Then $$ (x^2-y^2)^2k^6=(x+2\sqrt{xy}+y)^3. $$ We can see the expanded right hand side is of the form $m+n\sqrt{xy}$ for some integers $m,n$, and therefore $\sqrt{xy}$ is a rational number. It's not hard to see that this implies that $\sqrt{xy}$ is an integer, so $xy$ is a perfect square. Thus we can assume $x=la^2,y=lb^2$ where $(a,b)=1$ and $l$ is a positive integer. Plugging back to the original equation and raising both sides to sixth power we get $$ k^6=\frac{l^3(a + b)^6}{l^4(a^4 - b^4)^2} = \frac{(a + b)^4}{l(a-b)^2(a^2+b^2)^2}. $$ Now we see that $l$ is a square of a rational number, but since it $l$ is an integer, it must actually be a perfect square. This means $x$ and $y$ are both perfect squares.
So WLOG assume $x=a^2,y=b^2$ and the equation becomes $$ k^3=\frac{(a+b)^2}{(a-b)(a^2+b^2)}. $$ In particular $a^2+b^2$ divides $(a+b)^2=a^2+2ab+b^2$. Hence $a^2+b^2$ divides $2ab$, but since both values are positive, we have $a^2+b^2\leq 2ab$, i.e. $(a-b)^2\leq 0$ and thus $a=b$, impossible. So there is no solution in positive integers.
If we allow $y=0$ then the problem simplifies to $k^6x=1$ which leads to $x=1$.