The answer is $(p,q,r)=(5,-1,-1)$

This is just a straightforward vieta jumping.Assume that $a,b$ is such a pair and let $\frac{a^2+b^2+a+b+1}{ab}=k$then we have : $b^2+(1-ka)b+a^2+a+1=0$ let $b'$ be the other root of the equation.then $b+b'=ka-1$ which means that $b' \in \mathbb{Z}$.Also we have $bb'=a^2+a+1$ so we get $b'>=1$ and $b' \le a$ we can continue this recurtion until $a=b=1$.Since $b+b'=ka-1 \Rightarrow b=ka-b'-1$ we get $q=r=-1$ also considering three first elements in the recurtion namely $1,1,3$ we will get that $p=5$. Note:The solution is really badly written so please clarify it if you can.