How to do this inverse Z-transform? Is this correct?
A rational transfer function

$$H(z)=\frac{Y(z)}{X(z)}=\frac{\sum_{n=0}^{N}b[n]z^{-n}}{1+\sum_{n=1}^{N}a[n]z^{-n}}\tag{1}$$

corresponds to the following difference equation:

$$y[n]=b[0]x[n]+b[1]x[n-1]+\ldots +b[N]x[n-N]-\\-a[1]y[n-1]-\ldots -a[N]y[n-N]\tag{2}$$

So the current output sample $y[n]$ can be computed from the current input sample $x[n]$ and $N$ past input and output samples. $N$ is the order of the difference equation $(2)$.

In your example you have

$$H(z)=\frac{Y(z)}{X(z)}=-c\frac{1-z^{-1}}{1-cz^{-1}},\qquad c=\frac{1}{1+\frac{2RT}{m}}\tag{3}$$

and, consequently,

$$y[n]=c\big(-x[n]+x[n-1]+y[n-1]\big)\tag{4}$$