To find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the equation:

$$f(x+y) + f(x)f(y) = f(xy) + 2xy + 1$$

for all real numbers $x$ and $y$, we proceed as follows:

First, let's substitute specific values for $x$ and $y$ to gain insights into the form of $f$.

### Step 1: Substituting $x = 0$ and $y = 0$
$$f(0+0) + f(0)f(0) = f(0 \cdot 0) + 2 \cdot 0 \cdot 0 + 1 \implies f(0) + f(0)^2 = f(0) + 1$$

This implies $f(0)^2 = 1$, so $f(0) = 1$ or $f(0) = -1$.

### Case 1: $f(0) = 1$
Substituting $f(0) = 1$ into the original equation:
$$f(x+y) + f(x)f(y) = f(xy) + 2xy + 1$$

Substitute $y = 0$:
$$f(x+0) + f(x)f(0) = f(x \cdot 0) + 2x \cdot 0 + 1 \implies f(x) + f(x) = 1 + 1$$
$$2f(x) = 2 \implies f(x) = 1$$

This function does not satisfy every condition when plugged back into the original functional equation. Thus, $f(x) = 1$ is not a valid solution unless $x = 1$.

### Case 2: $f(0) = -1$
Substitute $f(0) = -1$ into the original equation:

$$f(x+y) + f(x)f(y) = f(xy) + 2xy + 1$$

Substitute $y = 0$:
$$f(x+0) + f(x)(-1) = f(0) + 2x \cdot 0 + 1 \implies f(x) - f(x) = -1 + 1$$
$$0 = 0$$

This does not provide new information, so let's substitute $y = 1$:

$$f(x+1) + f(x)f(1) = f(x) + 2x + 1$$

From here, if we assume $f(x) = ax^2 + bx + c$ or any polynomial form, we find consistent functions, try:

### Validation with Potential Functions

Substitute $f(x) = 2x - 1$:
$$f(x+y) = 2(x+y) - 1, \quad f(x)f(y) = (2x-1)(2y-1), \quad f(xy) = 2xy - 1$$
$$f(x+y) + f(x)f(y) = (2x + 2y - 1) + (4xy - 2x - 2y + 1) = 4xy$$
$$f(xy) + 2xy + 1 = 2xy - 1 + 2xy + 1 = 4xy$$

Thus, $f(x) = 2x - 1$ satisfies the equation.

Substitute $f(x) = x^2 - 1$:
$$f(x+y) = (x+y)^2 - 1, \quad f(x)f(y) = (x^2 - 1)(y^2 - 1), \quad f(xy) = x^2y^2 - 1$$
$$f(x+y) + f(x)f(y) = (x^2 + 2xy + y^2 - 1) + (x^2y^2 - x^2 - y^2 + 1) \eqref{(x^2y^2 - x^2 - y^2 + x^2 + 2xy + y^2)}$$
$$f(xy) + 2xy + 1 = x^2y^2 - 1 + 2xy + 1 = x^2y^2 + 2xy$$

Thus, $f(x) = x^2 - 1$ satisfies the equation.

Substitute $f(x) = -x - 1$:
$$f(x+y) = -(x+y) - 1, \quad f(x)f(y) = (-x-1)(-y-1), \quad f(xy) = -xy - 1$$
$$f(x+y) + f(x)f(y) = (-(x+y) - 1) + (xy + x + y + 1) = xy$$
$$f(xy) + 2xy + 1 = -xy - 1 + 2xy + 1 = xy$$

Thus, $f(x) = -x - 1$ satisfies the equation.

### Conclusion
The functions that satisfy the original functional equation are:

$$\boxed{f(x) = 2x - 1, \quad f(x) = x^2 - 1, \quad \text{and} \quad f(x) = -x - 1}$$