In general exponentiation is not commutative. However most people are aware of the pair 2 and 4 as a pair of unequal numbers for which exponentiation is commutative.

Show that there are an infinite number of pairs of unequal numbers for which exponentiation is commutative.

Go further and show that there are an infinite number of pairs of unequal rational numbers for which exponentiation is commutative.

Assume x is a multiple of y in the equation x^y=y^x. Then let y=v*x and substitute. After a bit of algebra a parametric solution emerges:
x=v^(1/(v-1)) and y=v^(v/(v-1))
v=2 or v=1/2 generates the classic pair of 2 and 4.

Assume v is rational. Then 1/(v-1) = p/q for some coprime integers p and q.
Then x=((p+q)/p)^(p/q) and y=((p+q)/p)^((p+q)/q).
If q is not equal to 1 then irrational roots pop back in, so make q=1.
Then x=((p+1)/p)^p and y=((p+1)/p)^(p+1) form a nontrivial rational solution to x^y=y^x for any positive integer.