If we have a reducible quadratic function
\begin{equation*} P(x)=a_1x^2+b_1x+c_1=(rx-x_1)(tx-x_2),~x_1,x_2,r,t\in\mathbb{Z}, \end{equation*}
does there exist another irreducible quadratic function $Q(x)=a_2x^2+b_2x+c_2$ such that for every natural $x$, there exists a natural $y$ such that $Q(x)=P(y)$? For example, is there an irreducible quadratic function $Q(x)$ such that whenever $x\in\mathbb{N}$, $Q(x)$ is equal to
\begin{equation*} P(y)=y^2-1=(y+1)(y-1) \end{equation*}
for some $y\in \mathbb{N}$?
EDIT: This does not answer the question that was asked --- see comments
The answer is no if you restrict to real-valued quadratics (which presumably we can because all complex-valued quadratics are reducible...). You can convince yourself of this by drawing some pictures of quadratics.
Take for example the reducible $P(x)=x^2-1=(x-1)(x+1)$.
Now $P(2)=+3$ and $P(0)=-1$.
Now suppose we have an irreducible $Q(x)$ and $y_1,\,y_2\in\mathbb{R}$ such that $Q(y_1)=P(2)=+3$ and $Q(y_2)=P(x_2)=-1$.
As all quadratics are continuous this implies that $Q$ has a root between $y_1$ and $y_2$ and hence by the Factor Theorem has a factor and is thus reducible.