The following problem seems somewhat abstract but it would be nice to get it (dis)proven. Assume we are given a set $\mathcal B$ of $n$ binary vectors $\boldsymbol b_1,\dots,\boldsymbol b_n$, each of dimension $2^n-3$, i.e., \begin{align} \boldsymbol b_i\in \{0,1\}^{2^n-3}, \ 1\le i\le n. \end{align} The assumption is now that there is no set $\mathcal B$ s.t. the following constraints are fulfilled: \begin{align} \langle \boldsymbol b_i,\boldsymbol b_i\rangle&\overset{!}{=}2^{n-1}-1, \ 1\le i\le n, \tag{1} \\ \langle \boldsymbol b_i,\boldsymbol b_j\rangle&\overset{!}{=}2^{n-2}-1, \ i\neq j. \tag{2} \end{align} In words, $(1)$ ensures that every vector has exactly $2^{n-1}-1$ ones and $(2)$ ensures that each vector pair has exactly $2^{n-2}-1$ ones in common.
For $n=2$ it is true, since for $\mathcal B=\{\boldsymbol b_1,\boldsymbol b_2\}$, $\boldsymbol b_i\in\{0,1\}$, we only have $1$ possibility, i.e., $$ \mathcal B =\{0,1\}, $$ which does not adhere $(1)$.
It is also true for $n=3$, since if $(1)$ is adhered, then every $\boldsymbol b_i$ is of dimension $5$ with $3$ ones. If we now take two vectors $\boldsymbol b_1$, $\boldsymbol b_2$ which adhere $(2)$, then the vector $\boldsymbol b_1 + \boldsymbol b_2$ has no zero entries. If any $\boldsymbol b_3$ adheres $(1)$ it has $3$ ones and thus has to have more than a single one in common with $\boldsymbol b_1$ or $\boldsymbol b_2$, which violates $(2)$.
Intuitively it seems to me that this principle is easily extendable to any $n>3$, either through induction or maybe just the pigeonhole principle, but I cannot figure out how. There may also be a nice counter example.
I found a counter example for n=4:
[{0, 1, 2, 3, 4, 5, 6}, {2, 3, 6, 7, 8, 10, 12}, {3, 4, 5, 7, 9, 10, 11}, {0, 1, 4, 7, 8, 9, 12}])
Where I only wrote the indices of the 'true' (or 1) entries for each vector. So, {0, 1, 2, 3, 4, 5, 6} could also be written as (1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0)
Cheers