Is $\mathcal{Z}\{4\delta[n-8]\delta[n-8]\} = 4z^{-16}$?
The notation is a little obscure but the answer is simple.

Since $\delta[n-8]^2$ can be represented as : $$\delta[n-8]^2 = \delta[n-8] \delta[n-8]$$ and since the multiplication of the impulses can be simplified as :
$$\delta[n-8] \delta[n-8] = \delta[n-8]$$ , then you can conlcude that the $\mathcal{z}$ transform of the signal $4 \delta[n-8]^2 $ will be:

$$ \mathcal{Z}\{ 4 \delta[n-8]^2 \} = 4 z^{-8} $$ whose region of convergence is all $z$ except $z = 0$.

There is a pole at $z = 0$ and a zero at $ z \to \infty $