Is there a set theory in which the reals are not a set but the natural numbers are?

Is there any known axiomatization of set theory in which the real numbers are not a set, but the natural numbers and other infinite sets do exist?

Such a set theory would have an Axiom of Infinity, but not an Axiom of Power Set. I know that Kripke-Platek set theory has no Axiom of Power Set, but it is not clear to me whether the real numbers exist as a set in this theory or not.

In the type of set theory I am envisioning, the real numbers would exist as a class, as would the class of all subsets of the natural numbers, and they could be equivalent to the class of all ordinals, for example.

One advantage of such a set theory would be that it could admit the Axiom of Choice, and the Well-Ordering Theorem for all sets, without having to admit the well-ordering of the reals or the Banach-Tarski paradox.

There is also Pocket set theory. In this set theory, infinite sets exists, and they are all equinumerous. So every infinite set is countable. But the reals exist as a proper class. This is a second-order theory in the sense that the objects are classes, and sets are classes which are elements of other classes.

In some sense this theory is somehow related to third-order arithmetic. There sets of natural numbers are still objects, and we can quantify over sets of sets of natural numbers (which in Pocket set theory make proper classes).

In $\mathsf{KP}$, there need not be a set of all real numbers. The canonical example of this is $L_{\omega_1^{CK}}$, the first level of Godel's constructible universe which satisfies $\mathsf{KP}$. There are reals with $L$-rank arbitrarily high below $\omega_1^{CK}$, so any set of reals in $L_{\omega_1^{CK}}$ is not the set of all reals in $L_{\omega_1^{CK}}$.

EDIT: Of course, $L_{\omega_1}$ also has this property, as does $L_\alpha$ for many countable $\alpha$s. However, if $L_\alpha$ is the Mostwoski collapse of an elementary substructure of $L_{\omega_2}$ (say), then $L_\alpha$ will think that there is a set of all reals, even if $\alpha$ is countable - basically, there are long "gaps" in the countable ordinals where no new reals enter $L$.