jmc

Let $p$ and $q$ be the two integers; then $p,q\in \{2,4,6,8,10,12,14\}$, giving $7\times 7=49$ total possible pairs $(p,q)$. The question asks for the number of different values of $pq+p+q$. Notice that by Simon's Favorite Factoring Trick, $$pq+p+q=(p+1)(q+1)-1,$$so it suffices to find the number of different possible values of $(p+1)(q+1)$. Here, $p+1,q+1\in \{3,5,7,9,11,13,15\}$.

There are $7$ pairs $(p,q)$ where $p+1$ is equal to $q+1$; by symmetry, half of the $42$ remaining pairs correspond to swapping the values of $p$ and $q$, leaving $42/2=21$ pairs $(p,q)$. Since most of the possible values of $p+1$ and $q+1$ are prime factors that do not divide into any of the other numbers, we note that most of the values of $(p+1)(q+1)$ will be distinct. The exception are the numbers divisible by $3$ and $5$: $p+1,q+1\in \{3,5,9,15\}$; then, if $(p+1,q+1)=(3,15)$ or $(5,9)$, then $(p+1)(q+1)=45$.

Hence, there are exactly $21-1+7=\boxed{27}$ distinct possible values of $pq+p+q$.