Note that $ S_{n+1}-S_n = 2^n $. Also note that $ S_n $ is a geometric series with a sum equal to $ 2^0\cdot\frac{1-2^n}{1-2} = 2^n-1 $. Using the Euclidean Algorithm, we obtain: \begin{align*} \text{gcd}(S_{n+1}, S_n) &= \text{gcd}(S_{n+1}-S_n, S_n) \\ &= \text{gcd}(2^n, 2^n-1) \\ &= \text{gcd}(2^n - (2^n-1), 2^n-1) \\ &= \text{gcd}(1, 2^n-1) \\ &= 1.\end{align*} Therefore, the greatest common divisor of two consecutive terms is always $ 1 $, so the largest possible value is $ \boxed{1} $.