My original problem was I wanted to find numbers where $13^a = 2^b$ but I realized this had a broader problem to it.
So the pretty obvious thing to first point out is that an even number to any power will still be even and an odd number to any power will still be odd. So if x or y is even and the other is odd then a and b are impossible. Disproving my idea above.
I assume the answer is that a and b are impossible for all x and y's but I'm having trouble putting it into a working proof and my mathematical vocab is not that strong. I'm also almost positive it has something to do with factorization ie. $x^4 = y^4 * z^4$ where $x=y*z$ and that all numbers can be broken up into prime factors but I'm still hindered by my lack of understanding. If anyone could explain how to find a and b or why it's impossible it would put my mind at ease.
For example, note that $9^3 = 3^6 = 27^2$ is possible. Actually, for $x, y\in\mathbb{N}$, there exist $a, b\in\mathbb{N}$ with $\gcd(a, b) = 1$ s.t $x^a=y^b$ if and only if there is $z\in\mathbb{N}$ s.t $x=z^b, y=z^a$. In this case the equation reduces to $z^{ab}=z^{ba}$, which is trivial.
This can be proved by prime factorization. Disregarding the trivial case $x=1$ or $y=1$, we can factorize $x, y$ as $$x = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r},\ \ \ \ \ y = q_1^{e_1}q_2^{e_2}\cdots q_s^{e_s}$$ where $p_i$'s are distinct primes and the same for $q_j$'s. Then the equation leads to $$p_1^{ak_1}p_2^{ak_2}\cdots p_r^{ak_r}=q_1^{be_1}q_2^{be_2}\cdots q_s^{be_s}$$ so that the uniqueness of prime factorization forces that $r=s$, each $p_i = q_j$ for some $j$, and $ak_i = be_j$. Then $a\mid e_j$ and $b\mid k_i$. Thus $$z = p_1^{k_1/b}p_2^{k_2/b}\cdots p_r^{k_r/b} = q_1^{e_1/a}q_2^{e_2/a}\cdots q_s^{e_s/a}$$ works.