jmc

Let $a,a+2$ be the two integers. We are given that $(a+2{)}^2-a^2=128$. Using the difference of squares factorization, the equation becomes $(a+2+a)(a+2-a)=128$. Simplifying and solving, we get: $$\begin{array}{rl}(2a+2)(2)& =128\\ \Rightarrow 2a+2& =64\\ \Rightarrow 2a& =62\\ \Rightarrow a& =31.\end{array}$$ Therefore the desired product is $a(a+2)=31\cdot 33=\boxed{1023}$.