I want to determine the set of equivalent values in exponential form based on a series of bases and powers. For example $2^k$, such that $2 \leq k \leq 100$ (ie, $2^{2-100}$) compared to $4^k$, $8^k$, $16^k$, $32^k$ and $64^k$. There will be a lot of equivalent values (e.g. $2^6 = 8^2$, $2^4 = 4^2$, $8^4 = 16^3$, etc.). I can brute force my way through this by identifying patterns, but I can't quite put this into a mathematical solution. I would appreciate any help in moving toward that solution.
I'm not a mathematician (obviously), so I apologize for my lack of familiarity with proper form and appropriate markup on this site.
Suppose $k$ is fixed (later on, we will sum from $k=1$ to $k=100$). We know that $$ \left(a^b\right)^c=a^{bc}=\left(a^c\right)^b $$ Now, we want all pairs $(x,y)\in\mathbb N^2$ with $x^y=2^k$. We know that $x$ is a power of $2$, so $x=2^z$ for some $z\in\mathbb N$. Now, we want to find all pairs $(z,y)\in\mathbb R^2$ such that $\left(2^z\right)^y=2^k$. Using the identity above, this simplifies to $yz=k$. Now, $y$ can be any divisor of $k$ and $z$ must be $\frac ky$. The number of divisors of a number is not really easy to compute, because you need the prime factorization of it. For $k\leq 100$, this can be done easily using a prime sieve.
Now, say $$ k=p_1^{q_1}\cdots p_n^{q_n} $$ then, the number of divisors of $k$ is $\prod_{i=1}^n (q_i+1)$.
I can expand this answer if you like. Please say so if needed.
EDIT:
If you want to find the actual divisors, I think the best way to get them is just by trial and error, because $k<100$.
For your example, $8^4=4^6$, you actually have $(2^3)^4=(2^2)^6$. This is true, because $k=3\cdot 4=2\cdot 6=12$. Other powers of two with the same value are corresponding with other divisor-pairs of $12$: $$1\cdot 12=2\cdot 6=3\cdot 4=4\cdot 3=6\cdot2=12\cdot 1$$ The powers of two are: $$ 2^{12}=4^6=8^4=64^2=(2^{12})^1=4096^1 $$