Answer: 16

Solution:

The value $m = 1$ fails, since the right side would be $2011$ while the left side is $1.$ For $m \ge 2,$ reduce mod $m - 1:$ every power of $m$ is $\equiv 1,$ so the equation forces $1 \equiv 2011 \pmod{m - 1},$ that is, $m - 1$ divides $2010.$

Conversely, suppose $2010 = (m - 1)n.$ Take $x_0 = n,$ let $m$ of the $x_k$ equal $0,$ and for each $r = 1, 2, \ldots, n - 1$ let $m - 1$ of the $x_k$ equal $r.$ This uses $m + (m - 1)(n - 1)$ $= n(m - 1) + 1$ $= 2011$ terms, and the sum telescopes: $$\begin{aligned} &m \\ &\quad {}+ (m - 1) \\ &\quad {}\cdot (m + m^2 + \cdots + m^{n-1}) \\ &\quad {}= m + (m^n - m) \\ &\quad {}= m^n = m^{x_0}. \end{aligned}$$

So the equation is solvable exactly when $m - 1$ divides $2010 = 2 \cdot 3 \cdot 5 \cdot 67,$ which has $2^4 = 16$ divisors. There are $16$ such $m.$