Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$

Following my answer here, we can deduce from the first equation that for some $a \neq 1$, we have
$$
x = a^{1/(a-1)}, \quad y = a^{a/(a-1)}
$$
Thus, the second equation becomes
$$
a^{2/(a-1)} = a^{3a/(a-1)} \implies\\
a = a^{3a/2} \implies\\
1 = a^{3a/2 -1} \implies\\
\left(\frac 32 a  - 1\right) \log a = 0
$$
Thus, we get a solution at $a = 2/3$, which is to say
$$
x = (2/3)^{-3} = \frac{27}{8}, \qquad y = (2/3)^{-2} = \frac{9}{4}
$$
