Find all positive integer solutions for $3^x-2^y=1$. 
Find all positive integer solutions for $3^x-2^y=1$.

Quickly we can find the solutions $(1,1)$ and $(2,3)$. Now the claim is that for $x \ge 2$ and $y \ge 3$ there are no solutions.
The equation can be expressed as $3^x=2^y+1$ from where we can get to $3^x-3 = 2^y-2 \implies 3(3^{x-1}-1)=2(2^{y-1}-1)$ now from this clearly $2$ is not a multiple of $3$ so it must be that $$2^{y-1}-1 \equiv 0 \pmod 3 \implies 2^{y-1} \equiv 1 \pmod 3$$ and the order of $2$ modulo $3$ is $2$ so $y-1$ is a multiple of two say $y-2=2t$.
Plugging this back to the lhs of the equation we get $$3(3^{x-1}-1)=2(2^{2t}-1)=2(2^2-1)(2^{2(t-1)}+ \dots +1) = 6(2^{2(t-1)}+ \dots +1)$$ now $3$ is not a multiple of $6$ so we must have that $$3^{x-1}-1\equiv 0 \pmod 6 \implies 3^{x-1} \equiv1 \pmod 6$$ but noticing that the only way this can happen is if $x-1=0$ i.e. $x=1$, but this contradicts our assumption that $x \ge2$ thus there are no solutions for $x \ge 2$.
Is this a correct argument? I think there are multiple ways to do the problem, but this seemed most natural to me.

We are going to assume there is a larger solution than $9-8=1$  and get a contradiction.
We have
$$ 3^u = 2^v + 1.  $$ Subtract $9$ from both sides, we have
$$ 3^u - 9 = 2^v - 8. $$
Taking $u = x + 2$ and $v = 3 + y$ gives $9 \cdot 3^x - 9 = 8 \cdot 2^y - 8,$ or
$$  9 (3^x - 1 ) = 8 (2^y - 1).  $$ We think this is only possible with $x=y=0,$ so we assume there is a solution with $x,y > 0$ and get a contradiction.
First we have $9 |(2^y - 1),$ or $$ 2^y \equiv 1 \pmod 9.$$ This tells us that $  6 | y. $ Meanwhile
$$ 2^6 - 1 = 63 = 3^2 \cdot 7. $$ Therefore $7 | (3^x - 1).$ In turn, we find $6 | x.$
Next
$$ 3^6 - 1 = 728 = 2^3 \cdot 7 \cdot 13. $$ Therefore $13 | (2^y - 1).$ In turn, we find $12 | y.$
Next
$$ 2^{12} - 1 = 4095 = 3^2 \cdot 5  \cdot 7 \cdot 13. $$ Therefore $5 | (3^x - 1).$ In turn, we find $4 | x.$
Finally, $3^x - 1$ is divisible by $3^4 - 1 = 80.$ In particular,
$3^x - 1$ is divisible by $2^4 = 16.$ However, this contradicts $2^y - 1 \neq 0$ and $y > 0.$
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