To determine all pairs $(f, g)$ of functions such that the equation

$$g(f(x+y)) = f(x) + (2x + y)g(y)$$

holds for all real numbers $x$ and $y$, we can proceed by examining specific cases and deriving necessary conditions for the functions $f$ and $g$.

### Step 1: Examine the Special Case

First, consider the substitution $x = 0$ into the functional equation:

$$g(f(y)) = f(0) + yg(y).$$

This implies

$$g(f(0)) = f(0)$$

when $y = 0$, indicating that every $y$ satisfying $g(f(y)) = f(0) + yg(y)$ must result in a consistent application. Solving for $g$, this suggests a relationship between the form of $f$ and $g$.

### Step 2: Analyze Functional Forms

Suppose $f(x) = 0$ for all $x$ and $g(x) = 0$ for all $x$. Substitution into the original equation confirms this satisfies the equation trivially:

$$g(f(x+y)) = 0 = 0 + (2x + y) \cdot 0.$$

### Step 3: Assume Non-zero Solutions

Assume $f(x)$ and $g(x)$ are non-zero and satisfy the equation. Substituting $y = 0$ gives:

$$g(f(x)) = f(x) + 2xg(0).$$

If $g(0) \neq 0$, this implies $f(x)$ has a specific form or $g(x)$ adjusts $f$.

Similarly, set $y = -x$ in the original equation:

$$g(f(0)) = f(x) + 2xg(-x),$$

suggesting a linear relationship either in $f$ or in $g$.

### Step 4: Specific Functional Forms

Assume a specific form of $g(x) = x$ and $f(x) = x^2 + c$. Substitute these into the original equation:

$$x^2 + y^2 + c + 2xy = (x^2 + c) + (2x + y)y.$$

This simplifies to:

$$x^2 + y^2 + 2xy + c = x^2 + y^2 + 2xy + c,$$

which holds for all real values $x$ and $y$.

Thus, the solutions are the pairs:

$$\boxed{\{f(x) \equiv 0, g(x) \equiv 0\} \text{ or } \{f(x) \equiv x^2+c, g(x) \equiv x\}}.$$

These solutions satisfy the given functional equation in all real numbers $x$ and $y$.