= Solution
Put $V=\mathbb EX_n^2$. For $c\geq0$, the process $(X_m+c)^2$ is a nonnegative <submartingale>, by the <conditional Jensen inequality>. A crossing of $x$ by $X$ implies a crossing of $(x+c)^2$ by this squared process. Since $\mathbb EX_n=\mathbb EX_0=0$, the <Doob maximal inequality> gives
$$
\mathbb P\left(\max_{m\leq n}X_m\geq x\right)
\leq\frac{\mathbb E(X_n+c)^2}{(x+c)^2}
=\frac{V+c^2}{(x+c)^2}.
$$
For $V>0$, <differentiation> of the last expression gives $2(cx-V)/(x+c)^3$, so its minimum over $c\geq0$ occurs at $c=V/x$. Substitution proves the <one-sided martingale maximal inequality>
$$
\boxed{\mathbb P\left(\max_{0\leq m\leq n}X_m\geq x\right)
\leq\frac{V}{V+x^2}.}
$$
If $V=0$, then $X_n=0$ <almost surely> and $X_m=\mathbb E[X_n\mid\mathcal F_m]=0$ for every $m\leq n$, so the same conclusion holds. The constant is sharp: at one step take $X_1=x$ with probability $V/(V+x^2)$ and $X_1=-V/x$ otherwise. This has <mean> zero, <variance> $V$, and achieves equality.
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