Let $S_k=X_1+X_2+\cdots+X_k$ for every $k\geqslant1$, then $x(n)=m(n)-m(n-1)$ is the probability that $S_k=n$ for at least one (hence exactly one) index $k\geqslant1$. Hence your goal is to compute, for each $n\geqslant1$,
$$
x(n)=\sum_{k\geqslant1}\mathrm P(S_k=n).
$$
Here is a further hint: each $x(n)$ is the coefficient of $u^n$ in the series
$$
t(u)=\sum_{k\geqslant1}\mathrm E(u^{S_k}).
$$
Now, you should be able to compute the generating functions $u\mapsto\mathrm E(u^{X_k})$ and $u\mapsto\mathrm E(u^{S_k})$, to deduce from these an expression of $t(u)$, and finally that $x(n)=p$ for every $n\geqslant1$.
Edit For every $u$ in $(0,1)$ and every $k\geqslant1$, $\mathrm E(u^{S_k})=\sum\limits_{n\geqslant1}\mathrm P(S_k=n)u^n$ by definition of the generating function. Summing this over $k$ and using the fact that every term is nonnegative hence the order of the summations does not change the result, one gets
$$
t(u)=\sum_{k\geqslant1}\sum_{n\geqslant1}\mathrm P(S_k=n)u^n=\sum_{n\geqslant1}u^n\sum_{k\geqslant1}\mathrm P(S_k=n)=\sum_{n\geqslant1}x(n)u^n.
$$
Hence, each $x(n)$ is the coefficient of $u^n$ in the series expansion of $t(u)$.
Now, here is an elementary computation: for every $X$ distributed like the random variables $X_k$, $\mathrm E(u^X)=g(u)$ with
$$
g(u)=\sum_{k\geqslant1}p(1-p)^{k-1}u^k=pu\sum_{k\geqslant0}((1-p)u)^k=\frac{pu}{1-(1-p)u}.
$$
Since $S_k$ is the sum of $k$ independent copies of $X$, $\mathrm E(u^{S_k})=g(u)^k$ for every $k\geqslant1$. Summing this over $k$, one gets
$$
t(u)=\sum_{k\geqslant1}g(u)^k=\frac{g(u)}{1-g(u)}.
$$
An elementary computation yields
$\frac{g(u)}{1-g(u)}=\frac{pu}{1-u}$ hence $t(u)=\sum\limits_{n\geqslant1}pu^n.
$
By identification, $x(n)=p$ for every $n\geqslant1$.