Why not use a complex number in the exponent in the z-transform? I am trying to comprehend how the z-transform has come to be similar but different from its continuous counterpart, namely the Laplace transform.

It seems to me that the most parallel and intuitive approach to the deduction of a generalized form of the DTFT would be to substitute $j\omega$ for $z = \sigma + j\omega$, i. e.:
$$
X(z) = \sum_{n = -\infty}^{+\infty} x[n]z^{-n} = \sum_{n = -\infty}^{+\infty} x[n]e^{-(\sigma + j\omega)n}
$$

Instead, we find normally that $z$ is a polar-form complex number which is the product of its magnitude and $e^{j\omega}$. I can guess that probably such representation leads to a rectangular representation of the z-transform, eventually rendering the analysis of poles and zeros impractical due to the periodic nature of the DTFT, but I am not sure. Could anyone enlighten me?
$z^{-1}$ is the unit delay operator.

$z$ is the unit advance operator.

$s$ is the derivative operator.

$1/s$ is the integration operator.

the utility of the Z and Laplace transforms is to solve (analyze, predict) constant coefficient differential and difference equations algebraically.