Does $\frac{z^2}{(z^2-1)^{\frac{1}{2}}}$ have a simple pole at $\infty$? If we define $w=\frac{1}{z}$ and Laurent expand the function about $w=0$ we have:
$$\frac{1}{w(1-w^2)^{\frac{1}{2}}}=\frac{1}{w}+\frac{1}{2}w+\frac{3}{8}w^3+...$$
This implies that there is a simple pole about $w=0$, because the inverse powers of $w$ only extend to $-1$.
However, is we write the expansion in terms of $z$:
$$\frac{z^2}{(z^2-1)^{\frac{1}{2}}}=z+\frac{1}{2z}+\frac{3}{8z^3}+...$$ 
I'm not sure how to read this. How does this series, when written in terms of $z$, imply there being a pole at $z=\infty$?
Furthermore, I am working through a problem that asks me to use this Laurent expansion to evaluate $\int_{C_{\infty}}\frac{z^2}{(z^2-1)^{\frac{1}{2}}}$, where $C_{\infty}$ is the circle at infinity. If my expansion is correct how would I go about solving this integral? Surely the integral is infinite? 

I think your error here is where you expanded the function. Notice that 
$$g(z)=\frac{z^2}{(z^2-1)^{\frac{1}{2}}}$$
has no singularity at zero, but it does have a singularity at $z=-1$ and $z=1$. You cannot use the Laurent series about $z=0$ of
$$f(z)=\frac{1}{({z^2-1})^{\frac{1}{2}}}$$
to evaluate something as $z\rightarrow\infty$. We know that the Laurent series for a function about a point $z_0$ converges everywhere inside a circle of radius equal to the distance from $z_0$ to the closest singularity to $z_0$. Thus the radius of convergence is $R=1$ for the Laurent series about $z=0$. For this reason, there is no series expansion that will work for $f(z)$ as $z\rightarrow\infty$ because the radius of convergence will always be capped by however far your $z_0$ is from $z=1$. With all of that said, there is no reason to use the Laurent series for $f(z)$ because $f(z)$ converges near infinity. We simply see that 
$$\lim_{z\rightarrow\infty}g(z)=\lim_{z\rightarrow\infty}\frac{z^2}{(z^2-1)^{\frac{1}{2}}}=\lim_{z\rightarrow\infty}z$$
which is equivalent to (if $w=z^{-1}$)
$$\lim_{w\rightarrow 0}\frac{1}{w}$$
so $g(z)$ has a pole of order $1$ as $z\rightarrow \infty$.
