Only two functions $f(x) \equiv 0$ or $f(x) \equiv 3$ are solutions to the equation. Indeed, let's put $c = f(0)$. By substituting $y = 0$, we obtain a quadratic equation
$$(f(x))^2 - 2f(x) - c = 0,$$
whence $f(x) = 1 - \sqrt{1+c}$ or $1 + \sqrt{1+c}$ for any $x \neq 0$.
Suppose that there are two non-zero numbers $a$ and $b$ of the same sign ($ab > 0$) such that $f(a) = 1 - \sqrt{1+c}$ and $f(b) = 1 + \sqrt{1+c}$. Substituting $x = a$ and $y = b$ into the original equation, we get
$$(f(a+b))^2 - f(a+b) - 2 = 0,$$
whence $f(a+b) = -1$ or $2$. Let's consider two cases.

1) Case $f(a+b) = -1$. Since $a+b \neq 0$ (the sum of the same sign as both terms), then $f(a+b) = 1 \pm \sqrt{1+c}$ and then necessarily $1 - \sqrt{1+c} = -1$. This means $c=3$ and $f(a) = f(a+b) = -1$. For $x=a$ and $y = a+b$ the original equation gives a quadratic equation
$$(f(2a+b))^2 = f(2a+b) - 2,$$
which has no solutions for $f(2a+b)$, which is impossible.

2) Case $f(a+b) = 2$. Since $a+b \neq 0$, then $f(a+b) = 1 \pm \sqrt{1+c}$ and so necessarily $1+\sqrt{1+c} = 2$. This means that $c=0$ and $f(b) = f(a+b) = 2$. For $x=a+b$ and $y=b$, the original equation gives a quadratic equation for $f(a+2b)$:
$$(f(a+2b))^2 = f(a+2b) + 4,$$
which has roots $\frac{1\pm\sqrt{17}}{2}$. However, also $a+2b \neq 0$ and $f(a+2b) = 1 \pm \sqrt{1+0} = 0$ or $2$. Contradiction.

Thus, on each of the open intervals $] -\infty, 0[$ and $]0, \infty[$ the function $f$ takes only one value $1 - \sqrt{1+c}$ or $1 + \sqrt{1+c}$. Therefore, if $x$ and $y$ are non-zero numbers of the same sign, then $f(x+y) = f(x) = f(y) = 1 \pm \sqrt{1+c}$ and
$$(1 \pm \sqrt{1+c})^2 = 3(1 \pm \sqrt{1+c}),$$
whence $1 \pm \sqrt{1+c} = 0$ or $3$. Since $1 - \sqrt{1+c} \le 1 \le 1 + \sqrt{1+c}$, two options are possible:
$$1 - \sqrt{1+c} = 0 \iff c = 0 \quad \text{or} \quad 1 + \sqrt{1+c} = 3 \iff c = 3.$$
Now let's substitute $y = -x$ into the original equation:
$$(f(0))^2 = f(0) + f(x) + f(-x) \iff f(x) + f(-x) = c^2 - c \text{ for any } x \neq 0.$$
If $c=0$ then on the entire set $\mathbb{R}$ the function $f$ can take at most two values $(0$ or $1+\sqrt{1+0} = 2)$. If $f(a) = 2$ for some $a \neq 0$, then $f(a) + f(-a) = 2 + f(-a) = 0^2 - 0 = 0$, which is impossible, since $f(-a) = 0$ or $2$. This means $f(x) \equiv 0$.
If $c=3$ then on the entire set $\mathbb{R}$ the function $f$ can take at most two values $(3$ or $1-\sqrt{1+3} = -1)$. If $f(b) = -1$ for some $b \neq 0$, then $f(b) + f(-b) = -1 + f(-b) = 3^2 - 3 = 6$, which is impossible, since $f(-b) = -1$ or $3$. So $f(x) \equiv 3$.
We can check that the functions $f(x) \equiv 0$ or $f(x) \equiv 3$ are suitable.