Can TRUE parameter free Zermelo prove Cantor's theorem?

Question

Let True Parameter-Free Z refer to the first order set theory having the following axioms

Extensionality: $\forall A,B (\forall x (x \in A \iff x \in B) \implies A=B)$

Foundation: $\forall X (\exists x \in X \implies \exists y\in X \forall z \in X (z \notin y)) $

Pairing: $\forall A,B \exists X \forall y (y\in X \iff y=A \lor y=B)$

Union: $\forall A \exists X \forall y (y\in X \iff \exists a \in A (y \in a))$

Power: $\forall A \exists X \forall y (y\in X \iff \forall z \in y (z \in A))$

Infinity: $\exists N \forall n (n \in N \iff \forall I ((0 \in I \wedge \forall i \in I (i \cup \{i\} \in I)) \implies n \in I))$

True parameter free Separation: If $\phi$ , $\delta$ are formulas having only the symbol "$y$" occuring free in them, and if $\phi(z)$ is the formula obtained from formula $\phi$ by merely replacing each occurrence of the symbol "$y$" in $\phi$ by the symbol "$z$", then:

$ \exists X \forall y (y \in X \iff\phi \wedge \delta \wedge \exists Z \forall z (z \in Z \iff\phi(z) )) $

is an axiom.

/Theory definition finished.

Now my questions are:

1. Is this theory equivalent to Zermelo set theory?
2. Can this theory prove Cantor's theorem of the uncountability of the power set of $N$?

OF note is that this theory is different from the known parameter free versions of Zermelo which have what is referred to as 'parameter free' versions of separation, however, those are not truly parameter free, in reality, ONE parameter is used which is the set upon which separation applies. Here even that is not allowed!