RMO 2015! #4

Seeing that the two prime factors on the R.H.S are $2,3$ we can safely assume that, $x=2^a\times3^b;y=2^c\times3^d$ where $a,b,c,d$ are non-negative integers.Squaring $x$ and cubing $y$ and adding their powers of $2$ and $3$ we get the following two equations, $2a+3c=12;2b+3d=12$ ,the first equation having the solutions $(a,c)\equiv(0,4),(3,2),(6,0)$ ,hence we have $3$ values of $a$ and similarly we have $3$ values of $b$ from the second equation.So,that is a total of $9$ combinations for the powers of $2,3)$ ,giving us $9$ values of $x$ and similarly $9$ corresponding values of $y$ ,but since the power of $x$ is $2$ we can even have $-$ of all the values that we just found hence we have a total of $18$ solutions!And done!