If $m \ge 11$, then $n = 2^m - 1 > 2013$. Since there is only one way a grasshopper jumps from point $1$ to point $2013$ by $2012$ steps, we see that $m \le 10$.

In the following, we show that the answer is $m = 10$. To show this, we will prove a stronger proposition by induction on $m$: for any $k \ge n = 2^m - 1$ and any $x, y \in P_n$, the number of ways the grasshopper jumps from point $x$ to point $y$ by $k$ steps is even.

If $m = 1$, the number of ways is $0$, where $0$ is even.

If $m = l$, the number of ways is even. Then, for $k \ge n = 2^{l+1} - 1$, there are three kinds of routes from point $x$ to point $y$ by $k$ steps. We show that the number of ways is even for each kind of route.

(1) The route does not pass point $2^l$. So points $x$ and $y$ both are on one side of point $2^l$. By the induction hypothesis, there are even routes.

(2) The route passes point $2^l$ just once.
Suppose that the grasshopper is at point $2^l$ at the $i$-th step.

$(i \in \{0, 1, \dots, k\}, i = 0 \text{ means } x = 2^l, i = k \text{ means } y = 2^l)$.
We show that, for any $i$, the number of routes is even.

Suppose that the route is $x, a_1, \dots, a_{i-1}, 2^l, a_{i+1}, \dots, a_{k-1}, y$. Divide it into two sub-routes: from point $x$ to point $a_{i-1}$ of $i-1$ steps and from point $a_{i+1}$ to point $y$ of $k-i-1$ steps (for $i=0$ or $k$, only one sub-route of $k-1$ steps).

If $i-1 < 2^l - 1$ and $k-i-1 < 2^l - 1$, then $k \le 2^{l+1} - 2$, which contradicts $k \ge n = 2^{l+1} - 1$. So, we must have $i-1 \ge 2^l - 1$ or $k-i-1 \ge 2^l - 1$. By the induction hypothesis, there are even ways for a sub-route. So, by the Multiplication Principle, the number of ways is even.

(3) The route passes point $2^l$ no less than two times.
Consider the sub-routes from $2^l$ to $2^l$, the number of ways is even, since we can consider the routes symmetric to $2^l$. So by the Multiplication Principle, the number of ways is even.

Summing up, the maximal $m$ is $10$.