Are there integer solutions to $9^x - 8^y = 1$?

This came up in proving non-regularity of a certain language (powers of 2 over the ternary alphabet). Any clue to the above equation could help me move forward.

Edit: Of course, $x = 1, y = 1$ is a solution. I am looking for non-trivial solutions.

Except $x=1$ and $y=1$ there aren't any.

We have that

$$3^{2x} - 1 = 8^y$$

i.e

$$ (3^x + 1)(3^x - 1) = 8^y$$

Thus we must have that

$$3^x + 1 = 2^m, 3^x - 1 = 2^n$$

Thus $$2^m - 2^n = 2$$

i.e $$ 2^n(2^{m-n} - 1) = 2$$

Thus $n=1$ and $m=2$.

The only solution is $x=y=1$. This is a special case of Catalan's conjecture which was proven by Mihailescu in 2002: the only solution in natural numbers of $x^a - y^b = 1$ with $a,b\gt 1$ is $x=3$, $a=2$, $y=2$, and $b=3$.

But as Moron points out, it can be deduced much more elementarily.

Equation $\rm\ 3^{2x}-2^{3y}=1\ $ is an instance of various special cases of Catalan's Conjecture.

First,$\ $ making the specialization $\rm\ \ \: z,\:p^n = 3^x,2^{3y}\ $ below yields $\rm\ x = 1 = y\ $ as desired.

LEMMA$\ \ $ $\rm z^2 - p^n = 1\ \ \Rightarrow\ \ z,\:p^n = \:3\ ,\:2^3\ $ or $\ 2,\:3\ $ for $\rm\ \ z,\:p\:,n\in \mathbb N,\ \ p\: $ prime

Proof $\rm\ \ \ (z+1)\:(z-1)\: =\: p^n\ \ \Rightarrow\ \ z+1 = p^{\:j},\ \ z-1 = p^k\ $ for some $\rm\ j,\:k\in \mathbb N$

$\rm\quad \:\Rightarrow\ \ \ \ 2\ =\ p^{\:j} - p^k\ =\ p^k\: (p^{\:j-k}-1) \ \Rightarrow\ p^k=2\ $ or $\rm\ p^k = 1 \ \Rightarrow\ \ldots$

Second, it's simply the special case $\rm\: X = 3^x,\ Y = 2^y\: $ of $\rm\ X^2 - Y^3 = 1\:,\: $ solved by Euler in 1738. Nowadays one can present this solution quite easily using elementary properties of $\rm\ \mathbb Z[\sqrt\[3\]{2}]\:$, e.g see p.44 of Metsankyla: Catalan's Conjecture: another old diophantine problem solved. See also this MO thread and this MO thread and Schoof: Catalan's Conjecture. Note also that Catalan equations are a special case of the theory of generalized Fermat (FLT) equations, e.g. see Darmon's exposition.