$x$ and $y$ are natural numbers $x+y+21=3xy$. Find maximum possible integral power of $6$ in $(xy)$!

Question

How should I solve this question?
I am getting three pairs $3,3 ; 11,1 ; 1,11$.

Can there be any other pairs and why?
I am not able to reason out that this are the only three pairs.

Answer 1

Hint:

$$1+63=(3x-1)(3y-1)$$

Now what are the positive divisors$(\equiv-1\pmod3)$ of $64?$

Answer 2

Since $3xy$ grows a lot faster than $x+y+21$, we can bound the possible solutions. Assume $x\leq y$, then

$$3xy =x+y+21 \leq 2y+21$$

$$(3x-2)y \leq 21$$

If $x\geq 4$, it forces $y$ to be less than $x$, contradicting our assumption. So $x$ can take only the values $1, 2,$ or $3$. You have all the pairs.