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!

Note that
$$2\cdot 3^x = p^y - 1 = (p - 1)(p^{y-1} + p^{y-2} + \ldots + p + 1) \tag{1}\label{eq1A}$$
With $p$ being odd and $p \gt 3$, then
$$p - 1 = 2(3^{a}) \; \; \to \; \; p = 2(3^{a}) + 1 \tag{2}\label{eq2A}$$
for some integer $a \ge 1$. Next, using the lifting-the-exponent (LTE) lemma, we have
$$x = \nu_3(p^y - 1) = \nu_3(p - 1) + \nu_3(y) = a + \nu_3(y) \tag{3}\label{eq3A}$$
Letting $\nu_3(y) = b$, there's then an integer $c \ge 1$ where $3 \nmid c$ and
$$y = c(3^b) \tag{4}\label{eq4A}$$
We next have
$$\begin{equation}\begin{aligned}
p^y & = 2^y(3^{ay}) + y(2(3^{a}))^{y-1} + \ldots + y(2(3^{a})) + 1 \\
p^y - 1 & \gt 2^y(3^{ay}) \\
2(3^x) & \gt 2^y(3^{ay}) \\
3^{x} & \gt 3^{ay} \\
3^{a+b} & \gt 3^{ac(3^b)} \\
a + b & \gt a(3^b)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
If $b = 0$, we get $a \gt a$, which is not true, so $b \gt 0$. Next, we have from \eqref{eq5A} that
$$a + b \gt a(3^b) \; \; \to \; \; b \gt a(3^b - 1) \ge 3^b - 1 \tag{6}\label{eq6A}$$
which is not possible. Thus, there's no solution to \eqref{eq1A} with $x \ge 2$, $y \ge 2$ and $p$ being a prime number (actually, no integer $p \gt 3$ works).
