find total integer solutions for $(x-2)(x-10)=3^y$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

How many integer solutions ($x$, $y$) are there of the equation $(x-2)(x-10)=3^y$?
(A)1 (B)2 (C)3 (D)4 (E)5

If let $y=0$, we had
$$x^2 - 12x + 20 = 1$$
$$x^2 - 12x + 19 = 0$$
no integer solution for x
let $y=1$, we had $x^2 - 12x + 17 = 0$, no integer solution too.
let $y=2$, we had $x^2 - 12x + 11 = 0$, we had $x = 1, 11$.
let $y=3$, we had $x^2 - 12x - 7 = 0$, no integer solution.
let $y=4$, we had $x^2 - 12x - 61 = 0$, no integer solution.
and going on....
is there any other efficient way to find it? "brute-forcing" it will wasting a lot of time.

By the Fundamental Theorem of Arithmetic, $x-2$ and $x-10$ must be powers of $3$, so we may write $$x-2= \pm 3^a $$ and $$x-10 = \pm 3^b.$$ Substracting the latter from the former we get $$8=\pm 3^a\mp3^b \\ 2^3 =3^a(\mp3^{b-a}\pm1)= 3^b (\pm3^{a-b} \mp1),$$ whence, again by FTA, $(a,b)= (0,2),( 2,0)$ and thus $(x,y)= (1,2),(11,2)$ are the only pairs of integer solutions.
