By time-shift and convolution property we can infer that:
$$ y(t)=x(t)*x(t) $$
$$ y(t-1)=x(t-1)*x(t-1) $$
$$ Z\{y(t-1)\}=z^{-1}Z\{y(t)\}=z^{-1}Z\{x(t)*x(t)\}=z^{-1}X(z)X(z) $$
But, if the signal y is a convolution of a signal whith a delayed itself, $$ y(t)=x(t)*x(t-1) $$ What would be the Z-transform of it in terms of X(z)? $$ Z\{y(t)\}=Z\{x(t)*x(t-1)\}=? $$ Thanks in advance.
There is an error in the beginning of your post when you substitute the $t - 1$ in $y(t - 1) = x(t - 1) * x(t - 1)$. This is due to the unfortunate abuse of notation regarding the convolution of functions $f * g$ as $f(t) * g(t)$ instead, but your solution is still correct.
Let’s use $\tilde{}$ above a function name for shifting by one in general. Then it turns out that $\widetilde{x * x} = x * \widetilde{x}$, so the $\mathcal{Z}$-transform will still be $z^{-1} X(z) \cdot X(z)$ in your second example.
Proof of the identity:
$$\begin{align} (x * \widetilde{x})(t) &= \sum_k x(k) \cdot \widetilde{x}(t - k) \\ &= \sum_k x(k) \cdot x(t - k - 1) \\ &= \sum_k x(k) \cdot x(\color{red}{t - 1} - k) \\ &= (x * x)(t - 1) \\ &= (\widetilde{x * x})(t). \end{align} $$