I note from the archive discussion of the case for integers, which I'm fine with.
For the positive reals, I have set up axes for x, y and z = x ^ y. The solution set is then graphed by a space curve in the surface, necessarily symmetrical about the plane y = x. I need the equation of the orthogonal projection of this curve onto the x-y plane. (All I have shown so far is that $x ^ y > y ^ x$ in the region right of the line y = e and above the line y = x.)
Let $0<x<y$. Then denote $$r = \dfrac{y}{x}>1.$$ Denote too $$ a = \log_{r}(x).$$ Then $x=r^a$, $y = r\cdot r^a = r^{a+1}$, and $$ \large{x^y = (r^a)^{r^{a+1}}=r^{a r^{a+1}}}; \\ \large{y^x} = (r^{a+1})^{r^a} = r^{(a+1)r^a}.\tag{1} $$ Since $x^y=y^x$, we have $$ ar^{a+1} = (a+1)r^a; $$ $$ \dfrac{a+1}{a} = \dfrac{r^{a+1}}{r^a} = r;\tag{2} $$ so we have $1$-parametric solution; returning to $x,y$: $$ \begin{array}{|c|} \hline x = r^a = \left(\frac{a+1}{a}\right)^a; \\ y = r^{a+1} = \left(\frac{a+1}{a}\right)^{a+1}; \\ \hline \end{array}\tag{3} $$ where $a>0$ (to satisfy condition $r>1$ from eq. $(2)$).
Then $$ \large{x^y = \left(\frac{a+1}{a}\right)^{\frac{(a+1)^{a+1}}{a^a}} = y^x.} $$
A few comfortable examples:
\begin{array}{ll} a=0.1: & (x,y)=(11^{0.1}, 11^{1.1}); \\ a=0.2: & (x,y)=(6^{0.2}, 6^{1.2}); \\ a=\frac{1}{3}: & (x,y)=(\sqrt[3]{4}, 4\sqrt[3]{4}); \\ a=0.4: & (x,y)=(3.5^{0.4}, 3.5^{1.4}); \\ a=0.5: & (x,y)=(\sqrt{3}, 3\sqrt{3}); \\ a=1: & (x,y)=(2,4); \quad \mbox{(famous example; no?)} \\ a=2: & (x,y)=(1.5^2, 1.5^3)=(2.25, 3.375); \\ a=4: & (x,y)=(1.25^4, 1.25^5) =(2.44140625, 3.0517578125); \\ a=5: & (x,y)=(1.2^5, 1.2^6) = (2.48832, 2.985984); \\ a=10: & (x,y) = (1.1^{10}, 1.1^{11}); \\ \ldots \end{array}
As for me, these few examples are rather nice:
$$2^4 = 4^2;$$
$$ \large{\left(\sqrt{3}\right)^{\sqrt{27}} = \left(\sqrt{27}\right)^\sqrt{3};} $$
$$ \large{\left(\sqrt[3]{4}\right)^{\sqrt[3]{256}} = \left(\sqrt[3]{256}\right)^\sqrt[3]{4};} $$
$$2.25^{3.375} = 3.375^{2.25};$$
$$2.48832^{2.985984} = 2.985984^{2.48832}.$$
