8. Let $m$ be a positive integer, $n=2^{m}-1$, and the set of $n$ points on the number line be $P_{n}=\{1,2, \cdots, n\}$.

A grasshopper jumps on these points, each step moving from one point to an adjacent point. Find the maximum value of $m$ such that for any $x, y \in P_{n}$, the number of ways to jump from point $x$ to point $y$ in 2012 steps (allowing intermediate visits to points $x, y$) is even.
8. When $m \geqslant 11$, $n=2^{m}-1>2013$.

Since there is only one way to jump from point 1 to point 2013 in 2012 steps, this is a contradiction, so $m \leqslant 10$.
We will now prove that $m=10$ satisfies the condition.
We use mathematical induction on $m$ to prove a stronger proposition: $\square$

For any $k \geqslant n=2^{m}-1$ and any $x, y \in P_{n}$, the number of ways to jump from point $x$ to point $y$ in $k$ steps is even.
When $m=1$, the number of ways to jump is necessarily 0, so the conclusion holds.
Assume that the conclusion holds for $m=l$.
For $k \geqslant n=2^{l+1}-1$, we divide the paths from point $x$ to point $y$ in $k$ steps into three categories, and we only need to prove that the number of paths in each case is even.
(1) The path never passes through point $2^{\prime}$.
In this case, points $x$ and $y$ are on the same side of point $2^{l}$, and by the induction hypothesis, the number of paths is even.
(2) The path passes through point $2^{l}$ exactly once.
Let the $i$-th step ($i \in\{0,1, \cdots, k\}$) be the step that reaches point $2^{l}$, where $i=0$ indicates that point $x$ is point $2^{l}$, and $i=k$ indicates that point $y$ is point $2^{l}$. We need to prove that for any $i$, the number of paths is even.
Let the path be
$x, a_{1}, \cdots, a_{i-1}, 2^{l}, a_{i+1}, \cdots, a_{k-1}, y$.
We divide this path into two sub-paths: from point $x$ to $a_{i-1}$, which is $i-1$ steps; and from point $a_{i+1}$ to $y$, which is $k-i-1$ steps (for $i=0$ or $k$, there is only one sub-path, which is $k-1$ steps).

Since $k \geqslant n=2^{l+1}-1$, if $i-1<2^{l}-1$ and $k-i-1<2^{l}-1$, then
$i-1 \leqslant 2^{l}-2$ and $k-i-1 \leqslant 2^{l}-2$.
Adding these, we get $k \leqslant 2^{l+1}-2$, which is a contradiction.
Therefore, either $i-1 \geqslant 2^{l}-1$ or $k-i-1 \geqslant 2^{l}-1$ must hold.

By the induction hypothesis, the number of paths for at least one of the sub-paths is even.

By the multiplication principle, the number of paths that reach point $2^{l}$ at the $i$-th step is even.
(3) The path passes through point $2^{l}$ at least twice.
In this case, we pair the paths between the first and second visits to point $2^{l}$ by reflecting them along point $2^{l}$, so the number of such paths is necessarily even.
By mathematical induction, the proposition is proved.
In summary, the maximum value of $m$ is 10.