We'll take this way too far

The important thing here is to notice that

$2^ 0 + 2^1 + 2^2 + \ldots + 2^{1336} = 2^{1337} -1 < 2^{1337}$

This means that if the rightmost square is $+$ , then $A$ will be positive irrespective of how other squares are filled. This is because, even in the worst case scenario, $A = 2^{1337} - 2^0 - 2^1 - 2^2 - \ldots - 2^{1336}$ is positive. Each of the $1336$ squares can be either $+$ or $-$ , so using the rule of product, it can be done in $2^{1336}$ ways.

Similarly, if the rightmost square was negative, then $A$ would always be negative, even if the first $1336$ squares were $+$ . In this case we have no solution.

We get $B = 2^{1336}$ , so $\log _{2} B = \boxed{1336}$ $_\square$