Let $b$ be an odd integer greater than $1$. Is it possible for the multiplicative orders of $2$ modulo two different powers of $b$ to be equal? The multiplicative orders of $2$ modulo powers of $b$ form a monotonically increasing sequence. Therefore the question is equivalent to the one whether it is possible for the multiplicative orders of $2$ modulo two consecutive powers of $b$ to be equal.
 A: Yes. Recall that an odd prime $p$ is called a Wieferich prime if $2^{p-1} \equiv 1 \bmod p^2$. Currently there are two known examples of Wieferich primes: $1093$ and $3511$. However, it is conjectured there are infinitely many.
If $p$ is an odd prime, then the multiplicative order of $2 \bmod p$ is equal to the multiplicative order of $2 \bmod p^2$ if and only if $p$ is a Wieferich prime, which answers your question in the positive. This is known and I am explaining it below.
Let us denote the order of $2 \bmod p^i$ by $d_i$ ($i=1,2$). Observe we always have (i) $d_1 \mid d_2$, (ii) $d_1 \mid p-1$ (Fermat's Little Theorem).
Claim: If $d_1=d_2$ then $p$ is Wieferich. Proof: We have $d_2 = d_1 \mid p-1$ and so $2^{p-1} \equiv 1 \bmod p^2$.
Claim: If $p$ is Wieferich then $d_1=d_2$. Proof: Since $p$ is Wieferich, we have $d_2 \mid p-1$. We also always have $d_2 \mid d_1 p$, by a lemma sometimes known as `lifting the exponent', proved by writing
$$ 2^{d_1 p} - 1 = \frac{2^{d_1 p}-1}{2^{d_1}-1} (2^{d_1}-1)$$
and noticing that both terms are divisible by $p$. Indeed, $2^{d_1} -1 \equiv 0 \bmod p$ by assumption. Moreover, $$\frac{2^{d_1 p}-1}{2^{d_1}-1} = \sum_{i=0}^{p-1} 2^{d_1 i} \equiv \sum_{i=0}^{p-1} 1^i = p \equiv 0 \bmod p.$$
The conditions $d_2 \mid d_1 p$ and $d_2 \mid p-1$ imply $d_2 \mid \gcd(d_1p, p-1) = d_1$. Since also $d_1 \mid d_2$, we are done.
