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 $S_{2m+1,m}=2^m.$

The celebrated $(23,12,7)-$Golay code would be impossible without the fact that $S_{23,3}=2^{11}.$ The Golay code and the binary Hamming codes 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 $\sum_1^m\binom{n}{i}(k-1)^i=k^j.$ 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.

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.

I think that is it.

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 $S_{2m+1,m}=2^m.$

The celebrated $(23,12,7)-$Golay code would be impossible without the fact that $S_{23,3}=2^{11}.$ If I'm not mistaken, the fact that the Golay code and the binary Hamming codes are the only perfect binary codes 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 $\sum_1^m\binom{n}{i}(k-1)^i=k^j.$ You were asking about $k=2.$ There is a perfect $(11,6,5)$ $3$-ary code which would not be possible except 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.

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 $S_{2m+1,m}=2^m.$

The celebrated $(23,12,7)-$Golay code would be impossible without the fact that $S_{23,3}=2^{11}.$ The Golay code and the binary Hamming codes 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 $\sum_1^m\binom{n}{i}(k-1)^i=k^j.$ 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.