Substitute $y = 2$ to get $2+f(x)f(2) = f(x)+f(2)+f(2x)$, which simplifies to $f(2x)-1 = 4(f(x)-1)$ for all reals $x$.
Now, suppose that $x_0 \neq 0$ is a zero of $f(x)-1$. Then $f(2x_0)-1 = 4(f(x_0)-1) = 4 \cdot 0 = 0$, and so, $2x_0$ is also a zero of $f(x)-1$. Continue this process indefinitely to get that $2^kx_0$ is a zero of $f(x)-1$ for all non-negative integers $k$. But $f(x)-1$ is a non-zero polynomial, and thus, can only have finitely many zeros, a contradiction.
Therefore, the only zero of $f(x) - 1$ is $0$, i.e. $f(x)-1 = Cx^n$ for some constant $C$ and some non-negative integer $n$.
Now, use the fact that $f(2x)-1 = 4(f(x)-1)$ to determine $n$, and then use the fact that $f(2) = 5$ to determine $C$.