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I don't know what the strategy should be, but for large enough $n$, you lose with probability $1$. At each step, the sum of the blocks increases by $2$ or $4$, so at least one of $4\times 2^{m^2}-4$ and $4\times2^{m^2}-2$ is hit. These numbers have at least $m^2$ $1$s in their binary expansions. If you write either as a sum of powers of $2$, you need at least $m^2$ terms, and if you only use $m^2$ then they must all be different. This means the board would be filled, and no merges would be possible.