Prefix sums of Pascal triangle = powers of two The circle division problem asks for the number of (bounded) regions obtained after choosing $n$ points in general position on a circle and then cutting along all segments connecting the points (cut along the circle too). The first few answers are $1, 2, 4, 8, 16, \mathbf{31}, \ldots$, and the general formula is $f(n) = {n \choose 4} + {n \choose 2} + 1$. The reason why the sequence starts with binary powers is that we can expand the answer as $\sum_{k = 0}^4 {n - 1 \choose k}$. We also have $f(10) = 256$ since the sum above is a half of a Pascal triangle row.
Denote $S_{n, m} = \sum_{k = 0}^m {n \choose k}$ the prefix sum of $n$-th row of the Pascal triangle. Obviously, $S_{n, 0}$, $S_{n, n}$, $S_{2n + 1, n}$, and $S_{2^t - 1, 1}$ are binary powers. Are there other non-trivial families of $S_{n, m}$ that are always binary powers? Can we characterize all such entries?
 A: I think that is it. But the question turns out to be open.
Your first example is the case $m=4$ of $S_{n,m}=2^m$ for $n \leq m$ while $S_{m+1,m}=2^{m+1}-1$ and also $S_{2m+1,m}=2^m.$ These could be considered to correspond to “perfect” codes with $1$ and $2$ code words.
The celebrated  $(23,12,7)-$Golay code would be impossible without the fact that $S_{23,3}=2^{11}.$ The Golay code and the $(2^{n-1},2^n-n-1,3)$-Hamming codes (thanks in part to $S_{2^n-1,1}$)  are the only perfect binary codes. I (incorrectly) thought that perhaps this follows from a proof that there are no other non-trivial cases of $S_{n,m}$ a power of $2.$ 
For a perfect $k$-ary code it would be necessary to have  a case of $\sum_0^m\binom{n}{i}(k-1)^i=k^j.$ You were asking about $k=2.$ There is a perfect $(11,6,5)$ $3$-ary code which would not be possible were it not the case that $\binom{11}0+2\binom{11}1+4\binom{11}2=3^5.$
There are no other perfect linear $k$-ary codes. I'm not sure if the proof stems from having no other coincidences as above, at least for $k$ a prime power. 
