Conversion from laplace transform to z-transform
Your question makes no sense. Z transform is performed on a discrete signal/series.

Since $H(s)$ is a continuous function, you can't just calculate a Z-transform of $H(s)$ without first sampling it, to make it discrete. Also, it doesn't make much sense to do a time->spectrum transform (such as a Z-transform) on a spectral representation ($H(s)$)

I'm assuming that when you write "$Z-Transform(H(s))$", what you really want to do is to convert $H(s)\to H(z)$, meaning to calculate the Z-transform of $h[nT]$, where $h[nT]$ is $h(t)$, sampled at intervals of $T$, and $h(t)$ is the inverse Laplace transform of $H(s)$.

If I'm correct in my assumption, the transformation you are seeking is known as "star transform", which would provide a transform function $H^{*}(s)$, in terms of $e^{sT}$, which may be easily converted to $H(z)$ by way of the substitution $z=e^{sT}$.

Edit - some elaboration on the conversion process: what you need to do is calculate $H^{*}(s)$ from $H(s)$ using one of the two relations described in "Relation to Laplace transform" in the Wikipedia article, then do the substitution $z=e^{sT}$, to get $H(z)$.