I claim the answer is $2^{2021}$. Note that $b$ is of form \[ b=1+\epsilon_1 2^1 + \epsilon_2 2^2 + \cdots + \epsilon_{2022}2^{2022},\quad\text{where}\quad \epsilon_i \in\{-1,1\}. \]Note that as $b>0$, we must have $\epsilon_{2022}=1$. Next I show any choice of $\epsilon_1,\dots,\epsilon_{2021}$ induces a distinct $b$. For this, consider the expansion, \[ b-2^{2022}-1 = \sum_{1\le k\le 2022}\epsilon_k 2^k \implies \frac12\left(b-2^{2022}-1 + \sum_{1\le k\le 2022}2^k\right) = \sum_{1\le k\le 2022}\zeta_k 2^k, \]for $\zeta_k\in\{0,1\}$. Since base-2 expansion is unique, our claim is proven.