If $f$ is holomorphic on a neighbourhood of $z_0$, the Laurent expansion of $f$ around $z_0$ will be the same as the Taylor expansion around $z_0$. (I.e. the principal part vill vanish.)
(Assuming you mean the Laurent series that converges on some $0 < |z-z_0| < r$. There are other possible interpretations.)
To finish on the way you started, let $\gamma$ be a circle with centre $0$ and radius $0 < \rho < R$. Then the standard estimate tells you
$$\lvert c_n\rvert = \frac{1}{2\pi}\left\lvert\int_{\lvert w\rvert = \rho} \frac{f(w)}{w^{n+1}}\,dw\right\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\rho^n}\,d\varphi \leqslant \rho^{-n}\max \{\lvert f(\zeta)\rvert : \lvert\zeta\rvert = \rho\}.$$
That gives you an upper bound on $\limsup\limits_{n\to\infty} \sqrt[n]{\lvert c_n\rvert}$ in terms of $\rho$. As $0 < \rho < R$ was arbitrary, you get the desired conclusion.
Another way to obtain the result is to use Cauchy's integral formula,
$$f(z) = \int_{\lvert w\rvert = \rho} \frac{f(w)}{w-z}\,dw,$$
for $\lvert z\rvert < \rho$, and expand $\dfrac{1}{w-z}$ into a geometric series, which you know converges uniformly on the circle $\lvert w\rvert= \rho$. Interchanging summation and integration then yields the convergence of the Taylor series in $\lvert z\rvert < \rho$. Let $\rho\to R$.
Best Answer
One also get at the radius of convergence like this:
Set
$\alpha = \dfrac{1 + i}{\sqrt{2}}; \tag{1}$
then
$\vert \alpha \vert = 1 \tag{2}$
and so
$\vert \alpha^{-1} \vert = 1 \tag{3}$
as well; we write
$f(z) = \dfrac{1}{1 + i - \sqrt{2}z} = \dfrac{1}{\sqrt{2}} \dfrac{1}{\alpha - z} = \dfrac{1}{\sqrt{2} \alpha} \dfrac{1}{1 - \alpha^{-1} z}; \tag{4}$
for $\vert z \vert < 1$ we have
$\vert \alpha^{-1} z \vert = \vert \alpha^{-1} \vert \vert z \vert < 1; \tag{5}$
thus the series
$\dfrac{1}{1 - \alpha^{-1} z} = 1 + \alpha^{-1}z + \alpha^{-2}z^2 + \ldots = \sum_0^\infty (\alpha^{-1} z)^n \tag{6}$
converges. Allowing $\vert z \vert = 1$, we may take $z = \alpha$ and not only is
$\dfrac{1}{(1 - \alpha^{-1} z)} = \dfrac{1}{(1 - \alpha^{-1} \alpha)} = \dfrac{1}{1 - 1} \tag{7}$
undefined, but
$\sum_0^\infty (\alpha^{-1} z)^n = \sum_0^\infty (\alpha^{-1} \alpha)^n = \sum_0^\infty 1^n = \infty; \tag{8}$
the series clearly explodes for $z = \alpha$. Since this series converges for $\vert z \vert < 1$ but not for $z = \alpha$, the radius of convergence is clearly $1$. The same holds for the series of $f(z)$, by (4).