Finding a rational root on this particular two variables polynomial Some context.
While working on a larger proof, I needed to show that a particular homogeneous system of polynomial equations had no rational solution except for the trivial one.
I have reduced this task to the following problem, and if the polynomial stated below has no rational root, it's over.
The question.

Let's consider the following polynomial of $\mathbb Q[X,Y]$:
$$P(X,Y)=Y^3(45X^2+18X-9)+Y^2(6X^2-15X+6)+Y(5X^4+15X^3-14X^2-3X-1)-X^5+3X^4+2X^3-X+1.$$
Does $P$ have a rational root?

Some remarks.

*

*I know that there is no general algorithm or method to answer such a question, but since this is a particular case, maybe there is an answer here.


*I think Faltings' theorem (though I don't understand it that well) shows that the rational roots of $P$ are in finite number.


*We can assume $Y\ne 0$, since there are no rational root of the form $(X,0)$. Thanks to a comment of Dietrich Burde, we can also assume $X\ne 0$ and $X\ne Y$.


*I thought of computing the roots of $P$ for $Y$ in terms of $X$, and finding conditions on $X$ (conditions of the form "$\sqrt{X^4-X^3+1}\in\mathbb Q$"), but it does not lead me anywhere.

If you are curious about it, this is what the curve defined by $P$ looks like:


Any enlightenment on this problem would be much appreciated.

A comment asked me what the original system was, so here it is:
$$\begin{cases}
-x_2^2+x_1x_3-x_4^2-x_2x_5-x_4x_5=0 \\
x_1^2-x_2^2+x_1x_4-3x_3x_4-x_2x_5-3x_3x_5=0 \\
x_1x_2-x_2x_3+x_2x_4-x_4x_5-x_5^2 =0 \\
3x_2x_3-x_2x_4-x_1x_5=0\\
3x_3^2-x_1x_4-x_4^2-x_2x_5-x_5^2=0
\end{cases}$$
for $x_1,\ldots,x_5\in\mathbb Q$.
 A: The OP's system in the $x_i$,
$$\begin{cases}
-x_2^2+x_1x_3-x_4^2-x_2x_5-x_4x_5=0 \\
x_1^2-x_2^2+x_1x_4-3x_3x_4-x_2x_5-3x_3x_5=0 \\
x_1x_2-x_2x_3+x_2x_4-x_4x_5-x_5^2 =0 \\
3x_2x_3-x_2x_4-x_1x_5=0\\
3x_3^2-x_1x_4-x_4^2-x_2x_5-x_5^2=0
\end{cases}$$
DOES have non-trivial integer solutions. Some small examples $x_1, x_2, x_3, x_4, x_5$ are,
$$18, 7, -19, 15, -28$$
$$133, 57, 46, -37, 75$$
$$380, 665, -97, 81, -651$$
and probably an infinite more.

Added details:
Given a system of $n$ equations in $n$ unknowns, we can generally resolve it into one equation in one unknown using resultants. 
The trick is to find the simplest equation to "cleave" the system. In the OP's case, it is the 4th one. Solve for its $x_5$, and we find,
$$x_5 =\frac{3 x_2 x_3 - x_2 x_4}{x_1}$$
Substitute this into the 1st and solve for $x_3$,
$$x_3 = \frac{x_1 x_2^2 - x_2^2 x_4 + x_1 x_4^2 - x_2 x_4^2}{x_1^2 - 3 x_2^2 - 3 x_2 x_4}$$
Substitute this $x_5, x_3$ and we find the remaining 3 equations become the SAME cubic in 3 variables $x_4, x_1, x_2$. Why this is so is a peculiarity of this system. 
It was then easy to use Mathematica to test small values such that the cubic in $x_4$ factors.
