What is the $\mathcal{Z}$-transform of a constant?
if you define the constant like this:
$$ x[n] = C ~~~,~~~\text{ for all } n $$

Then its $\mathcal{Z}$ transform
$$ X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} = \sum_{n=-\infty}^{\infty} C z^{-n} $$

does not converge for any value of $z$. Hence its $\mathcal{Z}$ transform does not exist.

Note that $\mathcal{Z}$ transform must be an analytic function with continuous derivatives of all orders unlike the DTFT which is not necesarily an analytic function and which would admit the use of impulses (or non-continuous functions) in its expressions.  Then this would allow the DTFT of the constant $x[n] =C$ to be $X(e^{j\omega}) =C 2\pi \delta(\omega)$. But as stated, impulses (which are not analytic functions) are not allowed to exist in $\mathcal{Z}$ transforms.

Had you defined the constant as the practical signal like
$$ y[n] = C u[n]$$ 
where $u[n] = \begin{cases} 1 &amp;,&amp; \text{ for } n \geq 0 \\ 0 &amp;,&amp; \text{ for } n &lt; 0\\ \end{cases} $ is the unit step function, then its $\mathcal{Z}$-Transform would exist:
$$ Y(z) = \sum_{n=-\infty}^{\infty} y[n]z^{-n} = \sum_{n=0}^{\infty} C z^{-n} = \frac{C}{ 1- z^{-1} } ~~~, ~~~ ROC~~ |z|&gt;1$$

The distinction is clear I suppose.