= Solution
For $m\geq1$, the increments have exponential moment
$$
\mathbb E2^{X_1}=\frac67\,2^{-1}+\frac17\,2^2=1.
$$
Thus $Z_n=2^{S_n}$ is an <exponential martingale of a random walk> relative to the natural <filtration>. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7)^{2m}>0$, independently of the preceding increments. The <geometric tail bound from a uniform escape probability> gives
$$
\mathbb P(T>2m\ell)\leq(1-q)^\ell,
$$
so $T$ is finite <almost surely> and has finite <expectation>.
At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\leq S_{n\wedge T}\leq m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded <optional stopping theorem> gives $\mathbb E Z_{n\wedge T}=Z_0=1$, and the <dominated convergence theorem> now yields
$$
\boxed{\mathbb E2^{S_T}=1.}
$$
Allowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.
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