The solutions for $x^x = z, z \in \mathbb{Z^+ - \{4\}}$ are irrational

Question

Suppose that the function $x^x = z, z \in \mathbb{Z^+ - \{4\}}$ has rational solutions. Then $x^x = (\frac{a}{b})^{{\frac{a}{b}}}$ where $a$ and $b$ are coprimes. Now let's look at the equation $(\frac{a}{b})^{{\frac{a}{b}}} = z$, clearly $(\frac{a}{b})^a = z^b$, so $\sqrt[a]{\frac{a^a}{b^a}} = \sqrt[a]{z^b}$. We conclude that $\frac{a}{b} = \sqrt[a]{z^b}$, this imples that both sides are rational, but the right side can only be rational if $z^{\frac{b}{a}}$ is rational, since $b$ and $a$ are coprimes, this cannot be true, so the assumption that the solutions are rational must be false.

This is my attempt to prove, it's clearly missing something. Does exist a more "technical" proof of the statement that every solution for this equation is irrational whenever $z$ is different of $4$?

Answer 1

$$ 27 = 3^3 $$ $$ 3125 = 5^5 $$

Answer 2

Your relation can be written $$ a^a=b^az^b $$ If $a=1$, then we conclude $a=b=z=1$.

Let's assume $a>1$. Then $a$ is divisible by a prime and we conclude $b=1$, because no prime dividing $b$ can divide $a$. Thus the relation becomes $a^a=z$, which is certainly possible: just take any positive integer $a\ne2$ and define $z=a^a$.

What is true is that the equation $x^x=z$ has no rational solution if $z$ is not of the form $a^a$ for an integer $a$.