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 $(x, y) = (3, 3)$.