Erdős problem 1119

Let $\mathfrak{m}$ be an infinite cardinal with $\aleph_0 < \mathfrak{m} < \mathfrak{c} = 2^{\aleph_0}$. Let $\{f_\alpha\}$ be a family of entire functions such that, for every $z_0 \in \mathbb{C}$, there are at most $\mathfrak{m}$ distinct values of $f_\alpha(z_0)$. Must $\{f_\alpha\}$ have cardinality at most $\mathfrak{m}$?

This is Problem 2.46 in [Ha74], where it is attributed to Erdős. The question is independent of ZFC, so the headline statement carries answer(sorry): it is neither provable nor refutable from the usual axioms of set theory.

The answer is yes if $\mathfrak{m}^+ < \mathfrak{c}$ (see erdos_1119.variants.easy_case), so the question reduces to the case $\mathfrak{m}^+ = \mathfrak{c}$, where it is undecidable: Kumar and Shelah [KuSh17] produced a model of $\mathfrak{c} = \aleph_2$ in which the answer is yes (with $\mathfrak{m} = \aleph_1$), while Schilhan and Weinert [ScWe24] produced a different model of $\mathfrak{c} = \aleph_2$ in which the answer is no.