Assume $p\mid ab+1, a+b$ for some prime $p$. Then, we have $(a+1)(b+1)= ab+1+a+b\equiv 0 \pmod p$. Thus, we have $a+1\equiv 0$ or $b+1\equiv 0$. WLOG, we assume $a\equiv -1$. Since, $a+b\equiv 0$, we must have $b\equiv 1$. Then, $b^2+a\equiv 0$. Therefore, $b^2+a=p$ must be satisfied. Then, $p>b^2\geq b\equiv 1$. Therefore, $b=1$, a contradiction.

assume $r\mid ab+1 $ and $r\mid a+b$ ( $r$ is prime) so $r\mid ab-a-b+1=(a-1)(b-1)$ ,WLOG $r\mid b-1$ so $r\mid b^2-1$ so $r\mid b^2+a+b-1$ then $r\mid b^2+a : prime$ implies $r=1$ ( that solves the problem) or $r=b^2+a$ now : $b^2+a\mid a+b$ that means $b$ is greater than or equal to $b^2$,contradiction.

Similar to post #2, except the finish is a little different. Assume for the sake of contradiction that $\gcd(ab+1,a+b) > 1$. Let $p$ be a prime which simultaneously divides both $a$ and $b$. If $p = 2$, then both $a$ and $b$ are odd, so $a^2 + b$ and $b^2 + a$ are both even and hence must equal $2$; this contradicts the fact that $a$ and $b$ are both greater than $1$. Now suppose $p > 2$. Then $p$ also divides \[ (ab+1) + (a+b) = (a+1)(b+1)\quad\text{and}\quad (ab+1) - (a+b) = (a-1)(b-1). \]Hence either $p\mid a+1$ or $p\mid b+1$, and similarly $p\mid a-1$ or $p\mid b-1$. The assumption that $p > 2$ implies that $p\mid a+1$ and $p\mid a-1$ cannot simultaneously hold. Thus, without loss of generality, let $p\mid a-1$ and $p\mid b+1$. Now observe that \[ a^2 + b \equiv 1^2 + (-1) \equiv 1 - 1\equiv 0\pmod p. \]Since $a^2 + b$ is prime, it must be equal to $p$. But then $\tfrac{a+b}{p} = \tfrac{a+b}{a^2+b}$ must be an integer; this cannot be the case as $a^2 > a$. Hence our original assumption was wrong and $\gcd(ab+1,a+b) = 1$.

different sol notice that $(a^2+b)(b^2+a)=ab(ab+1)+(a^3+b^3)=ab(ab+1)+(a+b)(\frac{a^3+b^3}{a+b})$ let $p=(a^2+b),q=(b^2+a)$, Assume for the sake of contradiction that $\gcd(ab+1,a+b) > 1$ then $\gcd(ab+1,a+b)=pt$ or $\gcd(ab+1,a+b)=qt$ then $a^2+b \mid a+b$ or $b^2+a\mid a+b$ ,now : $a^2+b\mid a+b$ that means $b$ is greater than or equal to $b^2$,contradiction.

generalization: Let$a,b>1,n$ - are naturals, and $a^2+nb,na+b^2$ are primes. Prove $(ab+n^2,a+b)=1$