Write as 5^x=2*3^y1
Checking small values, 2*3^y1 = 2, mod3, while 5^x alternates between 1 and 2, mod3.
Since 5^(2n)=1, mod3, we can tighten LHS to 5^(2x1).
Since 5 does not divide RHS unless y={1,5,9,13..} RHS is of form 2*3^(4n3)1.
So now 5^(2x1)=2*3^(4y3)1.
Since with the exception of 5 itself, 2*3^(4n3)1 = 8, mod9, RHS must = 8 mod 9, which is the case only if 5 is raised to a power of (6x3):
2*3^(4n3)1 = 5^(6x3)
Now LHS = {5,2,4}, mod 7, while RHS = 6, mod 7.
So {x=0,y=0} and {x=1,y=1} are the sole solutions.
