I wonder about the z transform for each of these sequences: $$ h_{n} = {1,2,3,4,5,6,...,\infty}$$
versus
$$x_{n} = {0,1,2,3,4,5,6,...,\infty}$$
where n = 0 to $\infty$ in both cases.
My understanding: $$H(z) = \frac{z}{(z-1)^{2}}$$
$$x_{n} = h(n+1) = zH(z) - zh(0) = \frac{z^{2}}{(z-1)^{2}} - z$$
Is this correct?
At first, we identify both sequences like tihis:
$$ \begin{align} x_{n} &= \underbrace{0,1,2,3,4,5,6,...,\infty}_{ramp}=n\\ h_{n} &= \underbrace{1,2,3,4,5,6,7,...,\infty}_{shifted\,\, ramp}=n+1\\ \end{align} $$
If we go through the Z-transform definition, we have, on your case, these two sums:
$$ \begin{align} X(z)&=\sum_{k=0}^{\infty}{kz^{-k}} & H(z)&=\sum_{k=0}^{\infty}{(k+1)z^{-k}} \end{align} $$
Looking at a Z-transform table we have:
$$X(z) = \frac{z}{(z-1)^{2}}$$
Then we can open up $H(z)$:
$$ \begin{align} H(z)&=\sum_{k=0}^{\infty}{(k+1)z^{-k}}\\ &=\sum_{k=0}^{\infty}{kz^{-k}}+\sum_{k=0}^{\infty}{z^{-k}}\\ &=X(z)+\sum_{k=0}^{\infty}{1^{k}z^{-k}}\\ &=\frac{z}{(z-1)^{2}}+\frac{z}{z-1}\\ &=\frac{z(z-1)+z(z-1)^{2}}{\left(z-1\right)^3}\\ &=\frac{z(z-1)(1+z-1)}{\left(z-1\right)^3}\\ &=\frac{z^2}{(z-1)^{2}} \end{align} $$
Which is exactly the same as using the time-shift property shown in the comments and your own question!