Solution:

Answer: $\frac{1}{2}$.

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
$$2 a^{2} + (2 x^{2} - 2 x + 1) a + (-x^{2} + x - 1) = 0$$
The quadratic formula produces the discriminant
$$(2 x^{2} - 2 x + 1)^{2} + 8(x^{2} - x + 1) = (2 x^{2} - 2 x + 3)^{2},$$
an identity that can be treated with the difference of squares, so that $a = \frac{-2 x^{2} + 2 x - 1 \pm (2 x^{2} - 2 x + 3)}{4} = \frac{1}{2}, -x^{2} + x - 1$. 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.