SOLVED
Let $f(n)$ be minimal such that every triangle-free graph $G$ with $n$ vertices and diameter $2$ contains a vertex with degree $\geq f(n)$.

What is the order of growth of $f(n)$? Does $f(n)/\sqrt{n}\to \infty$?

Asked by Erdős and Pach. The lower bound $f(n)\geq (1-o(1))\sqrt{n}$ follows from the fact that a graph with maximum degree $d$ and diameter $2$ has at most $1+d+d(d-1)=d^2+1$ many vertices.

Simonovits observed that the subsets of $[3m-1]$ of size $m$, two sets joined by edge if and only if they are disjoint, forms a triangle-free graph of diameter $2$ which is regular of degree $\binom{2m-1}{m}$. This construction proves that $f(n) \leq n^{(1+o(1))\frac{2}{3H(1/3)}}=n^{0.7182\cdots},$ where $H(x)$ is the binary entropy function. In [Er97b] Erdős encouraged the reader to try and find a better construction.

In this note Alon provides a simple construction that proves $f(n) \ll \sqrt{n\log n}$: take a triangle-free graph with independence number $\ll \sqrt{n\log n}$ (the existence of which is the lower bound in [165]) and add edges until it has diameter $2$; the neighbourhood of any set is an independent set and hence the maximum degree is still $\ll \sqrt{n\log n}$.

Hanson and Seyffarth [HaSe84] proved that $f(n)\leq (\sqrt{2}+o(1))\sqrt{n}$ using a Cayley graph on $\mathbb{Z}/n\mathbb{Z}$, with the generating set given by some symmetric complete sum-free set of size $\sim \sqrt{n}$. An alternative construction of such a complete sum-free set was given by Haviv and Levy [HaLe18].

Füredi and Seress [FuSe94] proved that $f(n)\leq (\frac{2}{\sqrt{3}}+o(1))\sqrt{n}$.

The precise asymptotics of $f(n)$ are unknown; Alon believes that the truth is $f(n)\sim \sqrt{n}$.

Additional thanks to: Noga Alon, Ishay Haviv