Prove that the following is not true or true. Prove that the following is not true or true. If $x,y,z \geq 1$ and they are all integers then $z^2=9xy+6y+6x+3$. If true, find $x,y,z$. If false, prove it.

I first tried some values but the fact that I am being asked to prove it if it is false, then I guess the above is false.

I tried some values like

$x=1,y=1$ then $z^2=24$ but that violates $z$ being an integer.

So if the above is false, how would i proceed in proving it.

EDIT: Sorry for all the confusion but I'm not trying to approach this problem by counterexamples. I want to know why does the equation $z^2=9xy+6y+6x+3$ not work? Why wont the left hand side ever provide a solution on the right hand side
So, there are two possible questions here: 

$1)$ As stated, your question states, prove or disprove that $\forall x,y,z\in \mathbb{Z}_{\ge 1}$ 
$$ z^2=9xy+6y+6x+3.$$
Here is the counterexample: consider $x=y=z=1$. This implies
$$ 1=24.$$
This is blatantly false, and so this can not hold for all $x,y,z,\in \mathbb{Z}_{\ge 1}$.

$2)$ The alternative question states, prove or disprove that $\exists x,y,z\in\mathbb{Z}_{\ge1}$ such that 
$$z^2=9xy+6y+6x+3. $$
Through software, or some sort of exhaustive calculation, we can see that $x=5,y=8\implies z=21$, a solution to the equation. So there exists such a $3-$tuple in $\mathbb{Z}_{\ge1}$ which solves the equation.
You haven't quite understood the question.

z^2 is defined as the result got by adding together 9xy + 6y + 6x +3 

you're being asked whether 9xy + 6y + 6x + 3 can be a perfect square,
when x and y are both positive integers.

Many integers are not squares; they want you to show 9xy + 6y  + 6x + 3 can be a square, presumably for just one set of values, though this seems unclear.

One approach might be to try to find x and y so that 3xy + 2y + 2x +1 is 3 x a square, such as 3, 12, 27, 48... maybe by factorising.

(x + 1)(y+1) = xy + y + x +1 umm..

or let's try x=y, so we get 9X^2 + 12x + 3 = 3(3x+1)(x+1).
Then say x+1 = 3M to allow for the 3
= 3(3M)(3x + 1)  to get a square, M (3x+1) must be square (where M=(x+1)/3)
so M( 3 (3M -1) + 1) has to be a square.
so M(9M -2) has to be square. So ... ummm...
Anyway, give it a go....