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Let $mathbb{A}$ denote the algebraic numbers.

Consider the set of numbers defined as follows:

$$S={2^{k_0}3^{k_1}5^{k_2}ldots;|;k_0, k_1, k_2ldotsinmathbb{Q}}$$
where $2,3,5dots$ are the prime numbers, and the number of primes factors must be finite (there must be finite $k_i neq0$). We can show that $Ssubseteqmathbb{A}$ by finding a polynomial with an arbitrary element from $S$ as a root:

Suppose $a$ is an arbitrary element from $S$,

For all rational exponents $k_0,k_1,k_2ldotsneq0$, let $m$ be the Least Common Multiple of their denominators. We know that $a^m$ is an integer because $m$ turns each quotient in the exponents into a whole number. Thus, the polynomial $x^m-a^m=0$ has $a$ as a root.

Next we assume that every combination of $k_0, k_1, k_2ldots$ produces a unique number in $S$, meaning that no two product of primes with distinct rational exponents may evaluate to the same number. We will attempt to show that the cardinality of $S$ is uncountable by diagonalization:

(From an overwhelming oversight, this method does not work because the number of primes was declared finite above.)

Suppose we have listed all numbers from $S$:
$$2^{frac{1}{3}} 3^{frac{2}{7}} 5^{frac{4}{3}} 7^{frac{1}{2}}ldots$$
$$2^{frac{2}{5}} 3^{frac{6}{5}} 5^{frac{1}{9}} 7^{frac{2}{3}}ldots$$
$$2^{frac{8}{3}} 3^{frac{3}{2}} 5^{frac{8}{7}} 7^{frac{1}{5}}ldots$$
$$vdots$$

We can produce a number in $S$ that is not in this list by constructing a new product where all of the exponents are the exponents along the diagonal plus $1$, e.g. $2^{frac{4}{3}}3^{frac{11}{5}}5^{frac{15}{7}}dots$ which will differ from every element in our list. $S$ is thus uncountable given our assumption.

However, the algebraic numbers are countable! And since $Ssubseteqmathbb{A}$, $S$ is indeed countable as well, which means that our assumption was false. There must exist some two prime products with distinct rational exponents that evaluate to the same number.

i.e.: $2^{s_0} 3^{s_1} 5^{s_2} ldots = 2^{t_0} 3^{t_1} 5^{t_2} ldots$ where $s_i$ and $t_i$ are rational and not always the same.

Question: Does the countability of the Algebraic numbers prove the existence of an equality of two distinct prime products with rational exponents?

Answer: No.

If so, what is a potential method for discovering these equalities?

I'm unconvinced that your diagonalization procedure produces a number not in the list that is also in your set $S$, because you have not demonstrated to me that only finitely many powers will be non-zero.

If you wish to hunt for collisions in $S$ there is also a more straight forward method you have already used; pick two representations in $S$, assume they are equal, and raise them both to an appropriate power (i.e. a number that will clear all denominators in every exponent). They will still be equal, and by unique factorization you will discover that they were in fact the same representation all along.

Does the countability of the Algebraic numbers prove the existence of an equality of two distinct prime products with rational exponents?

Not at all. Only finitely many exponents can be nonzero.