Factoring kind of problem
We know that $b$ is 256-bit number with
$$2^d - 1 = b\cdot s$$ and
$$z \equiv s \pmod{2^{128}}$$ use the modulo knowledge
$$z = s + k\cdot 2^{128}$$ for some $k \in \mathbb{Z}$ then
$$s = z - k\cdot 2^{128}$$ combine
$$2^d - 1 = b\cdot (z - k\cdot 2^{128})$$
$$2^d - 1 = b\cdot z - b \cdot k\cdot 2^{128}$$
take modulo $2^{128}$
$$ -1 \equiv b\cdot z  \pmod{2^{128}}$$
Therefore we learned 128-bit information about $s$ with simple arithmetic. It is not a good way to hide the private key.