Writing $a=x+2 y^{2}$, the given quickly becomes $4 y^{2} a+2 x^{2} a+a+x=x^{2}+1$. We can rewrite $4 y^{2} a$ for further reduction to $a(2 a-2 x)+2 x^{2} a+a+x=x^{2}+1$, or $$\begin{equation*} 2 a^{2}+\left(2 x^{2}-2 x+1\right) a+\left(-x^{2}+x-1\right)=0 \tag{*} \end{equation*}$$ The quadratic formula produces the discriminant $$\left(2 x^{2}-2 x+1\right)^{2}+8\left(x^{2}-x+1\right)=\left(2 x^{2}-2 x+3\right)^{2}$$ an identity that can be treated with the difference of squares, so that $a=\frac{-2 x^{2}+2 x-1 \pm\left(2 x^{2}-2 x+3\right)}{4}=$ $\frac{1}{2},-x^{2}+x-1$. Now $a$ was constructed from $x$ and $y$, so is not free. Indeed, the second expression flies in the face of the trivial inequality: $a=-x^{2}+x-1<-x^{2}+x \leq x+2 y^{2}=a$. On the other hand, $a=1 / 2$ is a bona fide solution to $\left(^{*}\right)$, which is identical to the original equation.