Any information on this sum I was wondering if we know what this sum converges to? How can we show it? It's just an odd looking sum I came across in some work.

It does converge.

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(2n)^{2^m} -1}$$

And/or it's alternating series.
Note that by Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

the series

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(2n)^{2^m} -1}$$

converges if and only if the following converges

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m}\frac{1}{(2n)^{m} -1}$$

which diverges since for m=1

$$ \sum_{n=1}^{\infty} \frac{1}{2n -1}$$

diverges.
Ok, lets assume it converges, I wrote a small code which takes few seconds to evaluate n=10000, and m=10000.
the partial sum is 

 0.5757556130311484


I will post the code here:

format long e
x=0;

for m=1:10000
    for n=1:10000
        x=x+1/((2*n)^(2^m)-1);
    end
end
x