It is easy to see that considering $x_0$ as a vector over $\mathbb{Z}_2$, we have $x_i = x_{i-1} + f(x_{i-1})$ where $f$ denotes the function which shifts all the elements of $f$ backwards by $1$. Then it is easy to see that: \[x_n = \sum_{i=0}^n f^i(x_0)\] utilizing the fact that $f$ is linear. Thus we have $x_m = x_0$ if and only if \[\sum_{i=1}^m f^i(x_0) = 0 \Longleftrightarrow \sum_{i=0}^{m-1} f^i(x_0) = 0\] Due to the fact $f^n(x) = x$ for any vector $x$ and that trivially the above holds when $n|m$ we can WLOG $m < n$ if $n \nmid m$. But then by keeping track of the first component of the vector we see that the LHS can never be equal to $0$ unless $m \ge n$ so we are done.