Unilateral Z transform
There is nothing wrong with your calculation. However, by definition, $x[n]=0$ for $n&lt;0$, because otherwise we couldn't have used the unilateral $\mathcal{Z}$-transform in the first place. So you get

$$\mathcal{Z}\big\{x[n-k]\big\}=x^{-k}X(z),\quad k&gt; 0\tag{1}$$

Eq. $(1)$ is valid for the unilateral as well as for the bilateral $\mathcal{Z}$-transform.

Things are different for a time advance. The unilateral $\mathcal{Z}$-transform of $x[n+k]$, $k&gt;0$, is

$$\begin{align}\mathcal{Z}\big\{x[n+k]\big\}&amp;=\sum_{n=0}^{\infty}x[n+k]z^{-n}\\&amp;=\sum_{n=k}^{\infty}x[n]z^{-(n-k)}\\&amp;=z^k\left(X(z)-\sum_{n=0}^{k-1}x[n]z^{-n}\right),\quad k&gt;0\tag{2}\end{align}$$

For the bilateral $\mathcal{Z}$-transform we simply have

$$\mathcal{Z}\big\{x[n+k]\big\}=z^kX(z)\tag{3}$$