Set $Q(x)=P(x)-10$. Now, $Q(2015)=2015$ and second condition implies $Q(x)^2+3=Q(x^2+3)$. Let $A=\{x|Q(x)=x\}$. So the condition impliess $x \in A \Rightarrow x^2+3 \in A$. Define $a_0=2015$ and $a_{i+1}=a_i^2+3>a_i$. Now we inductively have $Q(a_i)=a_i$. Since all $a_i$ are distinct, $R(x)=Q(x)-x$ has infinitely many roots implying $R(x)=0$ for all $x$ and hence $P(x)=x+10$ which is indeed a solution