How to find out the last pair

Well, I figured it this way :-

Clearly, one can deduce that both $x$ and $y$ must be powers of same prime number. Let $x=y^{p}$ where $p$ could be positive or negative or zero, for we don't know which among $x$ or $y$ is greater, of course it is possible to deduce that too beforehand but I supposed it this way.

Now, $x=y^{p}$ . Putting this in our equation gives us $y^{py^{2}}=y^{y^{p}}$ . Now, further simplifying leaves us with $p=y^{p-2}$ .

Now, we need to use logic and try putting different values for $p$ and $y$ . Clearly, $p$ could not be $5$ or more. One may now eisily put values and find that the only possible values of $p$ and $y$ are $(3,3), (4,2),(1,1)$ respectively, which leaves us with values of $x$ and $y$ as $(27,3),(16,2),(1,1)$ . Adding them gives us the answer as $\boxed {50}$