The Diophantine equation below comes from research in perfect binary codes:
$$\sum_{i=0}^e{n\choose i}=2^k.$$
Here we restrict ourselves to the case where $n$, $e$, $k$ are positive integers, $e>3$.
Question: I calculated all cases where $3<e<21$, $n\leq10^3$, and all the possible solutions seem to be $n=1,2,\cdots e,2e+1$.
My question is: is there a proof for this statement, or is there any counterexample for it? I have no idea how to prove whether there are any other solutions or not.
Thank you for your help.