= Solution
We prove that (i) implies (ii). Since $b$ is primitive modulo $p$, write
$$
b^{p-1}=1+cp.
$$
Being primitive modulo $p^2$ is equivalent to $p\nmid c$. The binomial theorem then gives inductively
$$
v_p\left(b^{p^{j}(p-1)}-1\right)=j+1.
$$
It follows that the order modulo $p^m$ is exactly
$$
p^{m-1}(p-1)=\varphi(p^m),
$$
so $b$ is primitive modulo every $p^m$. This is <lifting a primitive root to odd prime powers>. The implication (ii)$\Rightarrow$(i) is immediate by taking $m=2$.
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