To prove that (1) is equivalent to (under the condition that the left side is not less than the right side)
$$2 m \equiv a^{20}+b^{11}\left(\bmod 2^{2011}\right)$$

We first prove: there exist odd numbers $a, b$ satisfying (2).
Notice that, for any odd numbers $x, y$, we have
$$x^{11}-y^{11}=(x-y)\left(x^{10}+x^{9} y+\cdots+y^{10}\right),$$

The right side of the above equation, $x^{10}+x^{9} y+\cdots+y^{10}$, is the sum of 11 odd numbers, which must be odd. Therefore, $x^{11}-y^{11} \equiv 0\left(\bmod 2^{2011}\right) \Leftrightarrow x \equiv y\left(\bmod 2^{2011}\right)$. This indicates that, modulo $2^{2011}$, the numbers $1^{2011}$, $3^{2011}, \cdots, \left(2^{2011}-1\right)^{11}$ are a permutation of the numbers $1, 3, 5, \cdots, 2^{2011}-1$. Thus, there exists an odd number $b_{0}$ such that $b_{0}^{11} \equiv 2 m-1\left(\bmod 2^{2011}\right)$.

Now, take a sufficiently small negative odd number $b$ such that $b \equiv b_{0}\left(\bmod 2^{2011}\right)$ and $2 m-1-b^{11} \geqslant 0$, then $2 m-1-b^{11} \equiv 2 m-1-b_{0}^{11} \equiv 0\left(\bmod 2^{2011}\right)$. Therefore, let $(a, b, k) = \left(1, b, \frac{2 m-1-b^{11}}{2^{2011}}\right)$, which satisfies (1).

Thus, there exist $a, b, k$ that satisfy the conditions.