jmc

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: $$\begin{array}{rl}\text{gcd}(S_{n+1},S_n)& =\text{gcd}(S_{n+1}-S_n,S_n)\\ & =\text{gcd}(2^n,2^n-1)\\ & =\text{gcd}(2^n-(2^n-1),2^n-1)\\ & =\text{gcd}(1,2^n-1)\\ & =1.\end{array}$$Therefore, the greatest common divisor of two consecutive terms is always $1$, so the largest possible value is $\boxed{1}$.