Let $\rm P(0,0): f(2^x+2y)=2^yf(f(x))f(y)$. $\rm P(x,0): f(2^x)=f(f(x))f(0)\,\,(1)$. $\rm P(0,x): f(2x+1)=2^xf(f(0))f(x)\,\,(2)$ $\rm P(0,-1): f(-1)=\dfrac{1}{2}f(f(0))f(-1)$.We have two cases: Case 1: $\rm f(-1)\neq 0$,then$\boxed {\rm f(f(0))=2}$ so $\rm (2)\Leftrightarrow f(2x+1)=2^{x+1}f(x) \,\,\,(3)$.Also $\rm (1)\overset{x=0}{\Rightarrow} f(1)=2f(0)$. $\rm P(1,\dfrac{-1}{2}): f(1)=\dfrac{1}{\sqrt{2}}f(f(1))f(\dfrac{-1}{2})\Leftrightarrow 2f(0)=\dfrac{1}{\sqrt{2}}f(f(1))f(\dfrac{-1}{2})$. $\rm P(0,\dfrac{-1}{2}): f(0)=\dfrac{1}{\sqrt{2}}\cdot 2f(\dfrac{-1}{2})\Leftrightarrow 2f(0)=2\sqrt{2}f(\dfrac{-1}{2})$ If $\rm f(\dfrac{-1}{2})=0$ then $\rm f(0)=0\Rightarrow f(f(0))=0$ ,contradiction.So we have $ \boxed {\rm f(f(1))=4}$ $\rm P(1,x): f(2+2x)=2^x\cdot 4f(x)=2^{x+1}f(x)\cdot 2=2f(2x+1)\overset{2x+1\rightarrow x }{\Leftrightarrow } f(x+1)=2f(x)$. Let $\rm y\in \mathbb{N}$.We have $\rm f(2^x+2y)=2f(2^x+2y-1)=..=2^yf(2^x)$. So $\rm 2^yf(2^x)=2^yf(f(x))f(y)\Leftrightarrow f(2^x)=f(f(x))f(y)\overset {(1)}{\Leftrightarrow} f(f(x))f(y)=f(f(x))f(0),x\in \mathbb{R},y\in \mathbb{N}$. So $\rm f(f(x))=0,\forall x\in \mathbb{R}$ or $\rm f(y)=f(0),\forall y\in \mathbb{N}$. The first it is impossible because $\rm f(f(0))=2\neq 2$. The second it is also impossible because $\rm f(1)=f(0)\Rightarrow f(f(1))=f(f(0))\Leftrightarrow 4=2$. So there are no exist solutions in this case. Case 2 : $\rm f(-1)=0$ $\rm P(1,-1): f(0)=2^{-1}f(f(1))f(-1)=0\Leftrightarrow f(0)=0$. So $\rm f(f(0))=f(0)=0\Rightarrow f(2x+1)=0\overset{2x+1\rightarrow x}{\Leftrightarrow }f(x)=0,\forall x\in \mathbb{R}$ Therefore the only solution is $\rm f(x)=0$. This solution is wrong

Actually Case $1$ gives $f(x)=2^x$.

You're right, my solution has several errors.I hope the following solution is correct: Let $P(x,y):f (2^x + 2y) =2^y f ( f (x)) f (y) $ $ P(x,0): f(2^x)=f(f(x))f(0)$ $ P(0,x): f(2x+1)=2^xf(f(0))f(x)$ $ P(0,-1): f(-1)=\dfrac{1}{2}f(f(0))f(-1)$. If $f(0)=0$ then $ f(f(0))=f(0)=0\Rightarrow f(2x+1)=0\overset{2x+1\rightarrow x}{\Leftrightarrow }f(x)=0,\forall x\in \mathbb{R}$. Suppoce that $f(0)\neq 0$ We have two cases: Case 1 :$f(-1)=0$. $ P(1,-1): f(0)=2^{-1}f(f(1))f(-1)=0\Leftrightarrow f(0)=0$,contradiction. Case 2 :$f(-1)\neq 0\Rightarrow 2=f(f(0))$ $ P(1,x): f(2+2x)=2^x\cdot 4f(x)=2^{x+1}f(x)\cdot 2=2f(2x+1)\overset{2x+1\rightarrow x }{\Leftrightarrow } f(x+1)=2f(x)$ $f(f(x))=f(2^x)/f(0)$. So $P(log_2x,y): f(x+2y)=2^yf(y)f(x)/f(0),y\in R,x\in R_+$ If there exist $c\in R$ such that $f(c)=0$ then $y\rightarrow c\Rightarrow f(x)=0\forall x\geq c$ so since $f(x+1)=2f(x)$ we have that $f(0)=f(1)/2=f(2)/4=...=f(n)/2^n=0$ for $n$ big enough which is contradiction. For $y=-x$ we have that $f(-x)f(0)=2^{-x}f(-x)f(x)\Leftrightarrow f(x)=2^xf(0)\forall x\in R_+$ So $P(log_2x+1,-x): f(0)=2^{-x}f(-x)2^xf(0)/f(0)\Leftrightarrow f(-x)=2^{-x}f(0),x\in R_+$ so $f(x)=2^xf(0)\forall x\in R$ $f(f(0))=0\Leftrightarrow 2^{f(0)}f(0)=2$ and $f(f(1))=f(2)/f(0)\Leftrightarrow f(0)2^{f(1)}=2^2f(0)/f(0)\Leftrightarrow f(0)2^{2f(0)}=4=2f(0)2^{f(0)}\Leftrightarrow f(0)=1$ So the solutions are $\boxed{f(x)=0\forall x\in R},\boxed{f(x)=2^x,\forall x\in R}$