There will be infinitely many pairs $(x,y)$ of real numbers that satisfy the equation. If you are looking for integer solutions, that is another matter.
I would rewrite the equation as $9xy+42x+51y+213=0$, and then as
which turns into the attractive
Note that we did an analogue of completing the square. We get a hyperbola.
Now for integer solutions the analysis becomes quite simple. We take all ordered pairs $(s,t)$ of integers (both positive or both negative) such that $st=25$. There are only $6$ such pairs.
For some but not all of these pairs, it turns out that $x$ and $y$ are integers. Let's start. Look at $s=1, t=25$. No good, there is no integer $x$ such that $3x+17=1$. Look at $s=-1,t=-25$. That gives $3x+17=-1$, $3y+14=-25$, which has the integer solution $x=-6, y=-13$. Continue. There is not far to go!