Prove that $2\cdot 3^x +1= p^y$ has no solution Prove that the Diophantine equation of $3$ variables $(x,y,p)$
$$2\cdot 3^x +1= p^y$$
has no solution where $x,y\in\mathbb{N}_+$, $x\ge2, y\ge2$ and $p$ is a prime number.
I found that $y$ cannot be even, if $y =2k$ then $4| (p^k -1)(p^k+1) =2\cdot3^x$ which is a contradiction. But I cannot prove there is no solution for the case $y = 2k+1$.
I try to apply the technique in this answer (equation $7^x = 3 \cdot 2^y +1$) but it doesn't work as $p$ is not known.
Any help would be greatly appreciated!

We have, as the O.P. noted, the exponent of the prime $p$ is odd so
$$2\cdot3^x=(p-1)(p^{2y}+p^{2y-1}+\cdots+p+1)$$ the second factor having $y+1$ terms is odd so $2$ divides exactly $p-1=2m$,  with $m$ odd. It follows
$$3^x=m(p^{2y}+p^{2y-1}+\cdots+p+1)\Rightarrow m=3^a \text { and } \boxed{p=2\cdot 3^a+1} $$  then
$$3^{x-a}=(2\cdot 3^a+1)^{2y}+(2\cdot 3^a+1)^{2y-1}+\cdots+(2\cdot 3^a+1)+1$$ which implies $$3^{x-a}\equiv {1+1+\cdots+1}=2y+1\pmod{3^a}\iff3^{x-a}=2y+1+3^aX$$ then
$2y+1=3^ak$.
Back to the proposed equation we have $$2\cdot3^x+1=(2\cdot3^a+1)^{3^ak}$$ and similarly as above $$2\cdot3^x=2\cdot3^a[(2\cdot3^a)^{3^ak-1}+\cdots+(2\cdot3^a)+1]\Rightarrow 3^{x-a}\equiv 1\pmod{3^a}\iff3^{x-a}=1+3^aY$$ Thus
$$3^a(3^{x-2a}-Y)=1\Rightarrow 3^a=1\Rightarrow a=0\Rightarrow p=3$$
We are done.
