THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

The equality $2^x+{\log }_{3}x+x^2=2^y+{\log }_{3}y+y^2$ is readily obtained, and since the function $f:(0,+\infty )\to \mathbb{R},f(t)=2^t+{\log }_{3}t+t^2$ is increasing, hence one-to-one, we deduce that $x=y$.

Observe that $x=3$ is a solution for the equation $2^x+{\log }_{3}x=x^2$, while $1$, $2$, and $4$ are not. For $x\in \mathbb{N},x\ge 5$, we can easily show inductively that $2^x>x^2$. In conclusion, the given system has only one solution, namely $(x,y)=(3,3)$.