Parameterize the circle by $C(t) = z_0 + r e^{it} $ with $t\in (-\pi,\pi].$ Using the definition of a path integral, $ \int_{\gamma} f(z) dz = \int^b_a f(\gamma(t)) \gamma'(t) dt$
we have $ \int_{C} (z-z_0)^m dz = \int^{\pi}_{-\pi} r^m e^{imt} ire^{it} dt = ir^{m+1} \int^{\pi}_{-\pi} e^{i (m+1)t} dt.$
Now if $m+1\neq 0$ then the last integral is simply zero, seen either by direct evaluation or noting that the exponential goes through exactly $|m+1|$ periods. If $m+1=0$ then the integrand is $1$ and the value of the integral is thus $2\pi i.$
To summarize, $\int_C (z-z_0)^m \, dz = \left\{ \begin{array}{lr} 2\pi i \text{ if }m=-1 \\ 0 \text{ if } m\in \mathbb{Z}\setminus \{-1\} \end{array} \right.$