Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying the functional equation:

$$f(x^2 + f(y)) = y + (f(x))^2 \quad \text{for all } x, y \in \mathbb{R}.$$

Step 1: Evaluate the functional equation at specific points.

First, let's substitute $y = 0$ into the functional equation:

$$f(x^2 + f(0)) = (f(x))^2.$$

This equation will help us understand the behavior of $f$ for particular arguments.

Step 2: Consider $x = 0$ in the original equation:

$$f(f(y)) = y + (f(0))^2.$$

This implies that $f$ is bijective (since for any real $z$, there exists some $y = f(y')$ such that $f(z) = y'$ and $f(y') = z - (f(0))^2$).

Step 3: Substituting different values to study the parameter $f(0)$.

Suppose there exists some $a \in \mathbb{R}$ such that $f(a) = 0$. Then substituting $y = a$, we have

$$f(x^2) = a + (f(x))^2.$$

Since $f(f(a)) = a$, substituting $y = a$ into the equation of Step 2, we get:

$$f(0) = a + f(0)^2.$$

If $f(0) = 0$, it follows that $f(f(0)) = 0$, so $f(f(0)) = 0 = (f(0))^2$, consistent with $f(0) = 0$. Thus, we have $f(0) = 0$.

Step 4: Verify the potential solution $f(x) = x$.

Our goal is to verify $f(x) = x$. Substituting $f(x) = x$ into the original equation gives:

$$f(x^2 + y) = y + x^2,$$

which matches exactly with the right-hand side of the equation when $f(x) = x$.

Step 5: Conclude the proof.

We've shown that substituting $f(x) = x$ satisfies the original functional equation and that $f$ must be bijective, confirming that the only function $f$ that satisfies the equation is:

$$\boxed{f(x) = x}.$$

Thus, all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the given functional equation are in fact $f(x) = x$.