The conjecture is true, in fact the equation has no solution in distinct positive real numbers. To see this, let us write the equation in the more symmetric form
$$ x^y y^z z^x = x^z y^x z^y. \tag{$\ast$}$$
We get the same equation after interchanging $x$ and $y$, or $y$ and $z$, i.e., after permuting the variables arbitrarily. Hence we can assume without loss of generality that $x>y>z>0$. Then, with the notation $a:=x-y$ and $b:=y-z$, the original equation becomes
$$ (y+a)^b (y-b)^a = y^{a+b}, $$
where each factor and each exponent is positive. Equivalently,
$$ (1+a/y)^b (1-b/y)^a = 1, $$
where each factor and each exponent is positive. However, this is impossible, since
$$ (1+a/y)^b (1-b/y)^a < (e^{a/y})^b (e^{-b/y})^a = 1.$$

**Added on 22 January 2021.** Recently I posted the equation $(\ast)$ to a non-professional discussion board, and to my surprise two entirely new solutions arose. They are not mine, but I sketch them here as they are really nice and instructive. I will assume that $x,y,z>0$ are distinct and $(\ast)$ holds. I will derive a contradiction in two new ways.

**First new proof (sketch).** By assumption, $u:=y/x$ and $v:=z/x$ satisfy $u^{v-1}=v^{u-1}$. This contradicts (after some thought) the fact that the function $t\mapsto\frac{\ln t}{t-1}$ is strictly decreasing on the positive axis (the function is not defined at $t=1$, but it extends analytically there).

**Second new proof (sketch).** By assumption, the determinant
$$\begin{vmatrix}1&x&\ln x\\1&y&\ln y\\1&z&\ln z\\\end{vmatrix}$$
vanishes, hence its rows are linearly dependent. This contradicts (after some thought) the fact that the function $t\mapsto\ln t$ is strictly concave on the positive axis.

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