Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$

Consider the diophantine equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$ in positive integers $x_1\le x_2$, and $y_1\le y_2$. Each side of the equation is a product of two terms of a given binary recurrence, respectively. In this paper, we prove that the only solution to the title equation is $(x_1,x_2,y_1,y_2)=(1,2,1,1)$. The main novelty of our result is that we allow products of two terms on both sides.
10 pages, accepted for publication