Diophantine equation $p^{2n}+2=3^m$ Let $p$ be a prime and let $m$, $n$ be positive integers.
Consider the equation
$$p^{2n}+2=3^m$$
It is easy to see that $(p,m,n)=(5,3,1)$ is a solution. Are there any other solutions?

This isn't a full answer, but so far I've been able to show that $m$ has to be odd.
Because if $m$ is even, then $\frac{m}{2} \in \Bbb{N}$, and
$$p^{2n}+2=3^m$$
$$3^m-p^{2n} = 2$$
$$(3^\frac{m}{2}-p^n)(3^\frac{m}{2}+p^n) = 2$$
Since $\frac{m}{2}$ is a positive whole number, and $3^\frac{m}{2}+p^n > 3^\frac{m}{2}-p^n$, this can only happen when
$$\text{I.}\  3^\frac{m}{2}-p^n = 1 \ \ \text{  and } \\ \text{II.}\ 
 3^\frac{m}{2}+p^n = 2.$$
From the second equation:
$$p^n = 2 - 3^\frac{m}{2}$$
Putting this into the first equation:
$$3^{\frac{m}{2}}-2+3^{\frac{m}{2}}=1$$
$$2 \times 3^{\frac{m}{2}} = 3$$
Since the left side is even, and the right side is odd, this equation has no solution, so $m$ has to be odd.
