It is well-known that there are primes $p$ such that $2^{p-1} \equiv 1$ (mod $p^{2}$), a question which arises in connection with Fermat's Last Theorem. For such a prime $p,$ let $e$ be the smallest positive integer such that $p$ divides $2^{e}-1,$ and write (as we may) $p-1 = ed$ with $d$ an integer. Then we see easily that $2^{p-1}-1 \equiv d(2^{e}-1)$ (mod $p^{2}$). Certainly $d$ is not divisible by $p,$ so we must already have $2^{e} \equiv 1$ (mod $p^{2}$). Hence for such a prime $p,$ the multiplicative order of $2$ (mod $p$) is the same as the multiplicative order of $2$ mod $p^{2}.$
I see in the meantime that Francois Brunault has made a comment to similar effect, and that $1093$ is the smallest such prime