Let $A_i$ and $B_i$ denote the first and second move taken on the boxes marked $i$ and $i+1$. Let $i \le 6$. If Miha chooses to play $B_i$, then the number of tokens in the boxes marked $i$ and $i+1$ is reduced by $1$, the number of tokens in the box $i+2$ remains the same, and the number of tokens in the bag increases by $1$. If Miha chooses to play the sequence of moves $A_i$, $A_{i+1}$, $B_{i+1}$, then the number of tokens in the box $i$ decreases by $1$, the number of tokens in the box $i+1$ remains the same, and the number of tokens in the box $i+2$ and in the bag is increased by $1$. The end result is better in the second case since the only difference is that two boxes contain an extra token each. We conclude that the moves $B_i$ for $i \le 6$ are not profitable.

As for other moves, only $B_7$ increases the number of tokens in the bag, so Miha should make that move as many times as possible. Without loss of generality we may assume that this move is only taken at the very end. Before that, Miha uses $A_i$, $i \le 6$, to move the tokens into the seventh box which then contains $1 + 2 + 2^2 + \dots + 2^6 = 2^7 - 1 = 127$ tokens, while the final box contains $1$ token.

Let $x$ denote the number of tokens in the seventh box at a certain point in time, and let $y$ be the number of tokens in the last box at the same time. In order for Miha to make as many moves $B_7$ as possible, $\min\{x, y\}$ has to be as large as possible. Since $A_7$ decreases the number of tokens in the seventh box by $1$ and increases the number of tokens in the last box by $2$, this minimum will be the greatest when we have $x \le y$ for the first time, so when $y = x$.

After $k$ moves $A_7$ we have $x = 127 - k$ and $y = 1 + 2k$. Thus, we are looking for the smallest $k$ such that $127 - k \le 1 + 2k$, or $k \ge \frac{127-1}{3} = 42$. The smallest such $k$ is $k = 42$.

After $42$ moves of type $A_7$ Miha can use the $B_7$ moves to move $127 - 42 = 85$ tokens into the bag, and this is the greatest possible number of tokens that he can have in the bag at the end.