The bottom of this page provides a solution to the number of ordered factorizations of a positive integer with two prime factors: https://mathworld.wolfram.com/OrderedFactorization.html
i.e. if $n = p_1^{\alpha_1} p_2^{\alpha_2}$
Then numbers of ordered factorizations is $$H(n) = 2^{\alpha_1 + \alpha_2 - 1} \sum_{k=0}^{\alpha_2} {\alpha_1 \choose k}{\alpha_2 \choose k} 2^{-k}$$
And I am actually only interested in an easier version $m = \alpha_1 = \alpha_2$. In which case
$$G(p^m q^m) = 2^{2m - 1} \sum_{k=0}^{m} {m\choose k}{m\choose k} 2^{-k}$$
Is there a combinatoric argument to prove that result? I found the paper that proved it using generating function, but I'd be curious to see a combinatoric argument.
Also, how do I write $G(p^m q^m)$ as a generating function suppose we are able to infer this series through combinatoric argument?