For each irrational number $b$, does there exist an irrational number $a$ such that $a^b$ is rational?

What you want to prove is: given an irrational number $b$, then one of the following numbers : $$r^{\frac{1}{b}}\ \ \ \ \ \ r\in \Bbb Q $$ is irrational, this seems to be an attainable result for what we know nowadays about the transcendence of numbers, we have for example the following theorem:

Six Exponentials Theorem:Let $(x_1,x_2)$ and $(y_1,y_2,y_3)$ be two sets of complex numbers linearly independent over the rationals. Then at least one of $$e^{x_1y_1},e^{x_1y_2},e^{x_1y_3},e^{x_2y_1},e^{x_2y_2},e^{x_2y_3}$$ is transcendental.

Given an irrational number $x$, let $x_1=1,x_2=x$ and $y_1=\ln(p_1),y_2=\ln(p_2),y_3=\ln(p_3)$ for some primes $p_1,p_2,p_3$ hence using this theorem we have : at least one of: $$p_1,p_2,p_3,p_1^x,p_2^x,p_3^x $$ is irrational which gives us the following well known consequence:

Six Exponentials Theorem (special case). If $x$ is a real number such that $p_1^x$ , $p_2^x$ and $p_3^x$ are rational numbers for three distinct primes $p_1, p_2$ and $p_3$, then $x\in \Bbb Z$

If we use this theorem one $k^{\frac{1}{b}}$ of the numbers $2^{\frac{1}{b}},3^{\frac{1}{b}},5^{\frac{1}{b}}$ is irrational. and hence you can take $a=k^{\frac{1}{b}}$ and we have $a^b$ is an integer among $\{2,3,5\}$ which implies of course that it's rational.

Comment Maybe there is a very simpler answer which does not use this strong theorem.