Solving for $5^a+1 = 2\cdot3^b$ beyond $5+1=6$

Question

I am trying to solve $5^a +1=2\cdot3^b$ where $a,b >0$

I know that $a$ is odd and $b$ is of the form $4x+1$

Is there any solution besides $5+1=6$?

Answer 1

Working $\pmod 9$, we have that $5^3\equiv -1\pmod 9$, so every solution with $b\ge2$ needs that $a=3+6k=3(2k+1)$. But then $$5^{3(2k+1)}+1=(5^3+1)(\cdots)=126(\cdots)$$ And $126$ is a multiple of $7$, and the RHS is not.