Which extension of $\sf ZFC$ prove that $$ {\sf ZFC} \not \vdash \exists x \, ( \operatorname {CH}(x) \land x \neq \emptyset \land x \neq 1)$$

Where $\operatorname {CH}(x) \iff \neg \exists \kappa \, (|x| < \kappa < |P(x)|) $

In English: $\sf ZFC$ doesn't prove the continuum hypothesis of any set other than the empty set and the singleton of the empty set.

Which extension of $\sf ZFC$ prove that $$ {\sf ZFC} \not \vdash \exists x \, ( \operatorname {CH}(x) \land x \neq \emptyset \land x \neq 1)$$

Where $\operatorname {CH}(x) \iff \neg \exists \kappa \, (|x| < \kappa < |P(x)|) $

In English: $\sf ZFC$ doesn't prove the continuum hypothesis of any set other than the empty set and the singleton of the empty set.

I know that "$\sf ZFC + CH$ fails everywhere", can prove that, which is too strong. But I'm asking if this can be proved in a much less strong theory, some theory whose consistency strength is just a little above $\sf ZFC$.