jmc

Let $q(x)=p(x)-1.$ Then $p(x)=q(x)+1,$ so $$(q(x)+1)(q(y)+1)=q(x)+1+q(y)+1+q(xy)+1-2.$$Expanding, we get $$q(x)q(y)+q(x)+q(y)+1=q(x)+q(y)+q(xy)+1,$$so $q(xy)=q(x)q(y)$ for all real numbers $x$ and $y.$

Also, $q(2)=p(2)-1=4=2^2.$ Then $$\begin{array}{rl}q(2^2)& =q(2)q(2)=2^2\cdot 2^2=2^4,\\ q(2^3)& =q(2)q(2^2)=2^2\cdot 2^4=2^6,\\ q(2^4)& =q(2)q(2^3)=2^2\cdot 2^6=2^8,\end{array}$$and so on. Thus, $$q(2^n)=2^{2n}=(2^n{)}^2$$for all positive integers $n.$

Since $q(x)=x^2$ for infinitely many values of $x,$ by the Identity Theorem, $q(x)=x^2$ for all $x.$ Hence, $p(x)=q(x)+1=\boxed{x^2+1}.$