We denote by $m_{i,j}$ the number of marbles in Box $B_i$ after $j$ moves, and we define the associated sum $s_j$ by
$$s_j := \sum_{i=1}^{\infty} m_{i,j} 2^{i-1}.$$
We have $s_0 = 0$ and each $s_j - s_{j-1}$ is non-negative. In fact, $s_j - s_{j-1}$ equals:
$$\left. \begin{array}{c} 3 \\ 1+2^{M-1} \\ 0 \end{array} \right\} \quad \text{if the } j\text{'th move is} \quad \left. \begin{array}{c} A \\ B \\ C_k \end{array} \right\} \qquad (11)$$
It follows that $s_j > 1$ for all $j > 0$, and that a necessary condition to win after $j$ steps is that $s_j$ is a power of 2.
Suppose $M$ is even. Then, $1+2^{M-1}$ is a multiple of 3, so it follows inductively from (11) that $s_j$ is divisible by 3 for all $j$. We deduce that $s_j$ is never a power of 2. Thus, we cannot win the $M$-game if $M$ is even.
Suppose instead that $M$ is odd. Then $1 + 2^{M-1} \equiv 2 \pmod 3$. If we first do Move $B$ and then repeatedly do Move $A$, the numbers $s_j$ take on all values $v \ge v_0 := 1 + 2^{M-1}$ that are congruent to 2 (mod 3). We eventually reach a power of 2; in fact, using these moves, $s_j$ hits each odd power of 2 exceeding $v_0$.
Suppose that
$$s_j = \sum_{i=1}^{\infty} m_{i,j} 2^{i-1}$$
equals $2^P$ for some $P \in \mathbb{N}$. All our subsequent moves will be $C$-moves (meaning moves of the form $C_k$) and we continue until we can no longer perform a $C$-move. Each $C$-move lowers the total number of marbles by 1, so eventually no further $C$-moves are possible. Note though that the sum stays unchanged according to (11), so it is still $2^P$.
Suppose that after $J \ge j$ steps, we can go no further. This implies that $m_{i,J} \le 1$ for all $i$. Let $N$ be the largest value of $i$ such that $m_{i,J} = 1$, and suppose we are not in a winning position i.e. $m_{i,J} = 1$ for at least one value of $i < N$. Then
$$2^{N-1} < s_J = \sum_{i=1}^{N} m_{i,J} 2^{i-1} \le \sum_{i=1}^{N} 2^{i-1} = 2^N - 1.$$
This contradicts the assumption that $s_J$ is a power of 2. We conclude that the sequence of $C$-moves can end only when we reach a winning position.
We have shown that we can win if and only if $M$ is odd, and we have an algorithm for winning when it is possible.