Why does the Z transform represent a delay?

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I'm studying the Z-transform. I recently did by hand the Z transform of an discrete impulse delayed $\mathcal{z}\{\delta[n-k]\} = z^{-k}$

I get that this means that any signal can be represented as a linear combination of powers of $z^{-k}$. And this clearly has a direct link to the z transform of a diference for solving diference equations.

What I cannot grasp is why the use of an abritrary complex number Z in the convolution gives that particular result $\mathcal{z}\{\delta[n-k]\} = z^{-k}$ for a delay .