Through the method of convolution, I found that the inverse z transform of $$\frac{z^2}{(z-2)^2}$$ as $2^{n} \, (n+1)$.
But, when I try to use the same formula of convolution to find the inverse of $$\frac{z}{(z-2)^2}$$ I get the answer as $2^{n-1} \, (n+1)$ while the correct answer is $2^{n-1} \, n$.
Try a different method to see how it works. Consider the series $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$ which, by differentiation, $$\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} n \, x^{n-1}.$$ Now multiply the last equation by $x$ to obtain $$\frac{x}{(1-x)^2} = \sum_{n=0}^{\infty} n \, x^{n}.$$ Now, to be seen as the required z-transform set $x = 2/z$ to obtain the following result. $$\sum_{n=0}^{\infty} 2^{n} \, n \, z^{-n} = \frac{2}{z} \, \frac{1}{\left(1 - \frac{2}{z}\right)^2} = \frac{2 \, z}{(z-2)^2} $$ or $$\sum_{n=0}^{\infty} 2^{n-1} \, n \, z^{-n} = \frac{z}{(z-2)^2}.$$