Using partial fractions, for $1<|z|<2$, it results \begin{eqnarray*} \frac{(z+2)^2}{z^2-1} &=& 1+\frac{9/2}{z-1}-\frac{1/2}{z+1}\\&=& 1+\frac{9}{2z}\left(\frac{1}{1-\frac{1}{z}}\right)-\frac{1}{2z}\left(\frac{1}{1+\frac{1}{z}}\right)\\&=& 1+\frac{9}{2z}\sum_{n=0}^\infty z^{-n}-\frac{1}{2z}\sum_{n=0}^\infty (-z)^{-n}\\&=& 1+\frac{9}{2}\sum_{n=0}^\infty z^{-n-1}-\frac{1}{2}\sum_{n=0}^\infty(-1)^nz^{-n-1}.\qquad (\clubsuit) \end{eqnarray*} Now if $ n $ is even, then $ (- 1)^n=1$. Later $$ \frac{9}{2}\sum_{n = 0}^\infty z^{-n-1} - \frac{1}{2}\sum_{n = 0}^\infty (-1 )^nz^{- n-1} = 4\sum_{n = 0}^\infty z^{- n-1}. $$ But, if $ n $ is odd, then $ (- 1)^n = -1 $. So $$\frac{9}{2}\sum_{n = 0}^\infty z^{- n-1} -\frac{1}{2}\sum_{n = 0}^\infty (-1 )^nz^{- n-1} = 5\sum_{n = 0}^\infty z^{- n-1}. $$ For the above reasons, and noting that if $ n $ is odd, then $ -n-1 $ is even, and if $ n $ is even, then $ -n-1 $ is odd; from $ (\clubsuit) $ you have to $$\frac{(z+2)^2}{z^2-1}=1+4\sum_{m=0}^\infty z^{-2m-1}+5\sum_{m=0}^\infty z^{-2m}=1+\sum_{m=0}^\infty z^{-2m}\left(4z^{-1}+5\right).$$
This is my idea, but in the end I notice that this one is not in the form of a Laurent serial expansion in $1<|z|<2$.
Now my question point to point is, can I express the given function as a Laurent series in the indicated domain?