Hmm. The following argument (given using extended hints) disguises the basic bits about group theory (that you forbade). I don't know how many of these bits you have covered, so there is probably some overkill.
Let $m$ be any integer in the range $1\le mHow many elements are there in the set $[m]$? Well, we places no restrictions on $j$, so we need to answer, when might we have the congruence
$$
ma^j\equiv ma^{j'}\pmod x,
$$
where $0\le j
The next thing I want you to think about is: Is it possible that $[m_1]\cap[m_2]\neq\emptyset$? If $m_2a^{j_2}\equiv m_1a^{j_1}\pmod x$, then you can start cancelling the factors $a$. You should reach the conclusion that the intersection is non-empty, if and only if $m_2\in[m_1]$ and $m_1\in[m_2]$ in which case the two set coincide, i.e. $[m_1]=[m_2]$.
Now we are nearly done. Start with the set $\{1,2,\ldots,x-1\}$. Start casting out sets of the form $[m]$ for some $m$ still remaining. Conclude at each point that you are casting out exactly $2^{b+1}$ elements of this set (none of the newly casted were thrown out earlier). Eventually the set becomes empty, so conclude that the number of elements in the set must have been divisible by $2^{b+1}$ to begin with.