Let $P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}$. The problem then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_{2011}$ are nonnegative integers. We plug in $m = 1$ and see that $P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010$. Now, we can say that $P(m) = (m-1)Q(m) - 2010$ for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$, $(m-1)Q(m) = 2010$. Thus, if $P(m) = 0$, then $m-1 | 2010$ . Now, we need to show that for all $m-1 | 2010$, $m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}.$. We try with the first few $m$ that satisfy this. For $m = 2$, we see we can satisfy this if $x_0 = 2010$, $x_1 = 2009$, $x_2 = 2008$, $\cdots$ , $x_{2008} = 2$, $x_{2009} = 1$, $x_{2010} = 0$, $x_{2011} = 0$, because $2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots$ (based on the idea $2^n + 2^n = 2^{n+1}$, leading to a chain of substitutions of this kind) $= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}$. Thus $2$ is a possible value of $m$. For other values, for example $m = 3$, we can use the same strategy, with $x_{2011} = x_{2010} = x_{2009} = 0$, $x_{2008} = x_{2007} = 1$, $x_{2006} = x_{2005} = 2$, $\cdots$, $x_2 = x_1 = 1004$ and $x_0 = 1005$, because $3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004} = \cdots$ $=3^{1004} +3^{1004}+3^{1004} = 3^{1005}$. It's clearly seen we can use the same strategy for all $m-1 |2010$. We count all positive $m$ satisfying $m-1 |2010$, and see there are $\boxed{16}$