BALKAN-SL 2020 N4

Let $a_1=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1}=3a_n$ only for finitely many indicies $n$.

Proposed by Ilija Jovčevski, North Macedonia