Let $a = f(0)$. Setting $x = y = 0$ in the given equation gives $f(a) = a^2$. Setting $y = 0$ gives
$$f(x^2 + a) = (f(x))^2, \quad \forall x \in \mathbb{R}. \qquad (\circ)$$
Setting $x = a$ gives $f(a^2 + a) = (f(a))^2 = a^4$.
Assume that $a < 0$. Then there exists $b > 0$ such that $a = -b^2$. Setting $x = b$ in (\circ) gives $a = f(0) = f(b^2 + a) = (f(b))^2 \ge 0$, a contradiction. Hence $a \ge 0$.
Setting $x = \sqrt{a}$, $y = a$ in the given equation gives $f(a + a^2) = (f(\sqrt{a}) + a^2)^2$.
We have obtained $f(a + a^2) = (f(\sqrt{a}) + a^2)^2 = a^4$, i.e. $f(\sqrt{a}) (f(\sqrt{a}) + 2a^2) = 0$.
Assume $f(\sqrt{a}) = -2a^2$. Setting $x = \sqrt{\sqrt{a} + 2a^2}$, $y = \sqrt{a}$ gives
$$f(x^2 + f(y)) = f(\sqrt{a} + 2a^2 - 2a^2) = f(\sqrt{a}) = -2a^2 < 0.$$
But the given equation gives $f(x^2 + f(y)) = (f(x) + y^2)^2 \ge 0$, a contradiction. Hence $f(\sqrt{a}) = 0$.
Setting $x = 0$ in the given equation gives $f(f(y)) = (a + y^2)^2, \forall y \in \mathbb{R}$.
Setting $y = \sqrt{a}$ gives $a = f(0) = f(f(\sqrt{a})) = (a + a)^2 = 4a^2$.
We have two cases: $a = \frac{1}{4}$ or $a = 0$.
Assume $a = \frac{1}{4}$. Then $f(\frac{1}{2}) = 0$. Because $f(f(y)) = (a + y^2)^2$, we have $f(\frac{1}{4}) = \frac{1}{16}$. If $x$ and $y$ are real numbers such that $x^2 + f(y) = \frac{1}{2}$, the given equation gives $f(x) = -y^2$. This is true for, e.g., $x = \frac{\sqrt{7}}{4}$ and $y = \frac{1}{4}$ and we conclude that $f(\frac{\sqrt{7}}{4}) = -\frac{1}{16} < 0$. But setting $x = \sqrt{\frac{\sqrt{7}-1}{4}}$ in (\circ) gives $f(\frac{\sqrt{7}}{4}) \ge 0$, a contradiction.
Hence $f(0) = a = 0$ and we can simplify obtained identities to $f(x^2) = (f(x))^2$ and $f(f(y)) = y^4$.
From the first identity we have $f(x) \ge 0$ for all $x \ge 0$.
Assume there is $t > 0$ such that $f(t) < t^2$. Setting $x = \sqrt{t^2 - f(t)}$ and $y = t$ in the given equation gives
$$f(t^2) = \left( f(\sqrt{t^2 - f(t)}) + t^2 \right)^2 \ge t^4,$$
therefore $(f(t))^2 \ge t^4$ which is a contradiction with the assumption $f(t) < t^2$. Hence for all $x \ge 0$ holds $f(x) \ge x^2$.
This implies that $x^4 = f(f(x)) \ge (f(x))^2 \ge x^4$ for all $x > 0$, which is possible only if $f(x) = x^2$ for all $x \ge 0$.
Let $w > 0$. Setting $x = -w$ in $f(x^2) = (f(x))^2$ gives $f(-w) = w^2$ or $f(-w) = -w^2$. But if $f(-w) = -w^2$, setting $x = w$, $y = -w$ in the given equation gives $0 = f(0) = f(w^2 - w^2) = f(x^2 + f(y)) = (f(x) + y^2)^2 = 4w^4 > 0$, a contradiction.
Hence the only possible solution is $f(x) = x^2, \forall x \in \mathbb{R}$. We check directly that it really satisfies the given equation.