Example 8 Let $m$ be a fixed integer greater than 1, and the sequence $\left\{x_{n}\right\}$ is defined as follows:
$$
x_{i}=\left\{\begin{array}{ll}
2^{i}, & 0 \leqslant i \leqslant m-1 ; \\
\sum_{j=1}^{m} x_{i-j}, & i \geqslant m .
\end{array}\right.
$$

Find the maximum value of $k$ such that there are $k$ consecutive terms in the sequence that are divisible by $m$.
(44th IMO Shortlist)
【Analysis】Consider the sequence $\left\{y_{n}\right\}(n \in \mathbf{Z})$:
$$
\begin{array}{l}
y_{n}=y_{n-1}+y_{n-2}+\cdots+y_{n-m}, \\
y_{0}=1, y_{1}=2, \cdots, y_{m-1}=2^{m-1} .
\end{array}
$$

Then the sequence $\left\{x_{n}\right\}$ is exactly the part of the sequence $\left\{y_{n}\right\}$ with natural number indices.
Let $r_{n} \equiv y_{n}(\bmod m)\left(0 \leqslant r_{n} \leqslant m-1\right)$. Then $r_{n} \equiv r_{n-1}+r_{n-2}+\cdots+r_{n-m}(\bmod m)$, $r_{0}=1, r_{1} \equiv 2(\bmod m), \cdots, r_{m-1} \equiv 2^{m-1}(\bmod m)$.
Consider the ordered $m$-tuples $\left(r_{n}, r_{n+1}, \cdots, r_{n+m-1}\right)(n=1,2, \cdots)$ formed by consecutive $m$ terms of the sequence $\left\{r_{n}\right\}$. There are at most $m^{m}$ different such $m$-tuples. Therefore, there must exist two identical $m$-tuples $\left(r_{k}, r_{k+1}, \cdots, r_{k+m-1}\right)$ and $\left(r_{k+T}, r_{k+T+1}, \cdots, r_{k+T+m-1}\right)\left(T \in \mathbf{Z}_{+}\right)$, i.e., $r_{n}=r_{n+T}$ for $n=k, k+1, \cdots, k+m-1$.

Combining the $m$-order recurrence relation of $\left\{r_{n}\right\}$, by mathematical induction, we can prove:
When $n \geqslant k+m$, it always holds that $r_{n+T}=r_{n}$;
When $n \leqslant k-1$, it also holds.
Thus, $\left\{r_{i}\right\}$ is a purely periodic sequence with period $T$, and we can assume $T \geqslant m$ (if $T<m$, we can replace $T$ with $m T$).
By the definition of $\left\{r_{n}\right\}$, we know
$r_{-m}=0, r_{-(m-1)}=0, \cdots, r_{-2}=0, r_{-1}=1$.
Therefore, $r_{T-m}=r_{-m}=0$,
$r_{T-(m-1)}=r_{-(m-1)}=0, \cdots$,
$r_{T-2}=r_{-2}=0$.
Since $T \geqslant m$, then
$T-m, T-(m-1), \cdots, T-2 \in \mathbf{N}$.
Thus, there exist $m-1$ consecutive 0s in the part of $\left\{r_{n}\right\}$ with natural number indices.
Combining (1), it follows that there exist $m-1$ consecutive terms in the sequence $\left\{x_{n}\right\}$ that are divisible by $m$.

If there are $m$ consecutive 0s in the sequence $\left\{r_{n}\right\}$ modulo $m$, then by the $m$-order recurrence relation, it follows that for all $n \in \mathbf{Z}$, $r_{n}=0$, which is a contradiction.
Therefore, the maximum value of $k$ is $m-1$.
【Comment】The problem is transformed into considering the maximum number of consecutive 0s in the sequence modulo $m$, similar to Example 4, we can find the terms that meet the requirements by appropriately extending the indices of the initial terms of the sequence to negative integers.