Does this polynomial have a rational value which is the square of a rational number?

I have the following polynomial:

$$P(x,y,z):=9y^2z^2-30x^2z+90xyz+54yz-270x+81\in\mathbb Q[x].$$

It came up in a larger proof, and I would need in order to complete the proof to prove the following result:

Does there exist $$(x,y,z,r)\in\mathbb Q^4$$ such that $$x\ne 0$$ and

$$P(x,y,z)=r^2.$$

We can reformulate the problem in the following way:

Does the algebraic variety defined by

$$9Y^2Z^2-30X^2Z+90XYZ+54YZ-270X+81-T^2$$

have a rational point with $$X\ne 0$$?

I have no idea how to tackle this problem, I have looked up several articles, but nothing seems to apply to this particular question.

Any hints or references would be greatly appreciated.

• @TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience. – E. Joseph Nov 29 '18 at 10:54

Two obvious solutions are $$P(0,0,0)=(\pm9)^2$$.
To find more solutions, plugging in $$z=0$$ yields $$P(x,y,0)=-270x+81,$$ which is a square for $$x=\frac{81-t^2}{270}$$ for any choice of $$t\in\Bbb{Q}$$, and any choice of $$y\in\Bbb{Q}$$.