Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solutions $$(1,1,1),(1,2,2),(1,5,11)$$I already know $(2^x-1)(3^y-1)=z^2$ has no solution. See: P.G.Walsh December 2006 [On Diophantine equations of the form] paper but there is a factor of $2$ that seems complicated, and I didn't know anyone had studied this before. If so, please help me with the article or link, thanks.