Simple and elegant

Taking $x^2 = y^3 \Rightarrow y = x^{\frac{2}{3}},$ substituting this value into other equation will yield:

$x^y = y^x \Rightarrow x^{x^{\frac{2}{3}}} = (x^{\frac{2}{3}})^x;$

or $x^{\frac{2}{3}} = \frac{2x}{3} \Rightarrow x^2 = \frac{8x^3}{27};$

or $8x^3 - 27x^2 = 0 \Rightarrow x^{2}(8x - 27) = 0 \Rightarrow x = 0, \frac{27}{8}$ and $y = 0, (\frac{27}{8})^{\frac{2}{3}} = \frac{9}{4}.$

Hence the solutions are $(x,y) = (0, 0); (\frac{27}{8}, \frac{9}{4}).$ But $0^{0}$ is indeterminate, which only leaves $\boxed{x = \frac{27}{8}, y = \frac{9}{4}}.$