For each $x\in T$, denote $B_x=\{y\in T:d(x,y)\leq 3\}$, namely, the Hamming ball centered at $x$ with radius $3$. Note that, for every distinct $x,y\in S$, we have $B_x\cap B_y=\varnothing$. Indeed, if $B_x\cap B_y\supseteq \{z\}$ for some $z$, then both $x,y$ are within Hamming distance $3$ to $z$, and both are members of $S$, a contradiction. Furthermore, we also have, $\bigcup_{x\in S}B_x=T$, since for every $z\in T$, there is an $x'\in S$ such that $d(x',z)\leq 3$, thus, $z\in B_{x'}$. Now, one can obtain simply $|B_x|=\binom{n}{3}+\binom{n}{2}+\binom{n}{1}+1$. Using this, we have, $$ |T|=2^{2k-1}=2^k\cdot\left(\binom{n}{3}+\binom{n}{2}+\binom{n}{1}+1\right), \quad\text{where}\quad n=2k-1. $$Solving this for $k$, one immediately obtains that $k=12$, and therefore $n=23$ is the only solution. Edit. This is apparently from the 1989 Longlist.