I want to build an uncountable well-ordered set within the theory $Z^\textbf-$. So, I take $A=\omega$ (exists by infinity axiom) and define $$W:=\{(X,R)\in\mathcal{P}(A)\times\mathcal{P}(A\times A):\langle X,R\rangle\ \text{is a well-ordered set}\}$$
With this I consider the set $T:=W/\cong$ (where $\cong$ is isomorphism relation). Note that $W$ and $T$ are sets by power set axiom. So I define for each equivalence class $[x],[y]\in T$ the order
$$ [x]\leq_T[y]\Leftrightarrow \text{type}(x, R_x)\leq \text{type}(y,R_y) $$
Here, $R_x$ means the order of $x$ that make it belongs to $[x]$.
So, $T$ is well ordered by $\leq_T$ and it is uncountable.
Question: I don't pretty sure if I can build $\leq_ T$ without replacement axiom. Another doubt is, Can I take the $R_x$ without AC?
Hartogs' theorem is provable without Replacement. The trick is to note that the isomorphism with ordinals is really unnecessary.
Instead you want to just look at well-orders modulo order isomorphisms. In fact, one can look at $\{X\subseteq\mathcal P(A)\mid (X,\subsetneq)\text{ is well-ordered}\}/\cong$, or in other words, look at the set of chains of subsets of $A$ which are well-ordered by $\subseteq$, modulo the order-isomorphism relation.
It is not hard to show that this set is both well-ordered, and does not embed into $A$. So if $A=\omega$, the resulting order type is uncountable by definition.
Note that Replacement was never used here. We never mapped each well-order to its order type as a von Neumann ordinal. We just looked at sets of sets of chains of subsets, with a bunch of Separation.