Infinitely many $n\neq m$ with $\binom{2n}{n}$, $\binom{2m}{m}$ having the same prime divisors? (Erdős #730)

Are there infinitely many pairs of integers $n\neq m$ such that $\binom{2n}{n}$ and $\binom{2m}{m}$ have the same set of prime divisors? (Only the SET of primes dividing each central binomial coefficient matters, not multiplicities. Note that by Kummer's theorem a prime $p$ divides $\binom{2n}{n}$ if and only if adding $n+n$ in base $p$ produces at least one carry, so membership of each prime is decided by base-$p$ digit conditions.)

Acceptance. FULLY RESOLVES: a proof that infinitely many such pairs exist (e.g. by exhibiting an explicit infinite family with proof, or a density/counting argument), or a proof that only finitely many exist. Machine-checkable (Lean/Coq) proof preferred — the formal-conjectures statement can host it — else a complete written proof; finitely many verified examples do NOT close the problem. ADVANCES: an extension of the verified enumeration of A129515 to a record search bound, with code, the new confirmed terms, and per-pair machine-checkable witnesses (for each claimed pair, the lists of primes with the Kummer carry check); a first pair of shape $(n,n+k)$ for a value of $k$ with no previously known pair, or a run of four or more consecutive integers sharing one support (strictly extending the triple stated in the background); or a proven structural result (e.g. infinitude conditional on a stated standard hypothesis, or a positive-density statement for coinciding supports). Deliver the proof file, or the search code + bound + witness certificates.