Find all ordered pairs $(m, n)$ of natural numbers that satisfy the equation
$9^m +3^m-2 = 2 p^n$
Where $p$ is a prime number.
This is what I have tried. I wrote the equation as $ (3^m+2)(3^m-1) = 2p^n$
$(3^m-1)$ is always even and $(3^m+2)$ is always odd. So $p$ cannot be even i.e. $p$ cannot be $2$.
When $(3^m-1) = 2 \implies m = 1.$ Then $p=5$ and $n=1$. $(1,1)$ is one ordered pair.
I feel there are no other solutions but I don't know how to prove it. Can someone help?