According to this lemma:
$$ (2^a - 1, 2^b - 1) = 2^{(a,b)}-1 $$
This means if \((a,b)=1\), then:
$$ (2^a - 1, 2^b - 1) = 2^1 - 1 = 1 $$
Conversely, if \((2^a - 1, 2^b - 1) = 1\), then \(2^{(a,b)}-1 = 1\), which means \(2^{(a,b)} = 2\). Therefore, \((a,b)=1\).