Seventeenth Stars of Mathematics Competition

Suppose, if possible, that $(a^n + a + 1)(b^n + b + 1)$ is a square for all but finitely many positive integers $n$. Then
$$ab \equiv \frac{(a^{p+1} + a + 1)(b^{p+1} + b + 1)}{(a^{p-2} + a + 1)(b^{p-2} + b + 1)}$$
and
$$(a + 2)(b + 2) \equiv (a^{p-1} + a + 1)(b^{p-1} + b + 1) \pmod{p}$$
are both quadratic residues modulo all but finitely many primes $p$. Consequently, $ab$ and $(a+2)(b+2)$ are both squares. (It is a fact that a positive integer which is a quadratic residue modulo all but finitely many primes is itself a square.)
Let $b-a = q$, a prime, and let $d = \gcd(a, b)$. Clearly, $d=1$ or $d=q$. If $d=q$, then $a = (c-1)q$ and $b = cq$ for some integer $c > 1$. Hence $c(c-1)q^2 = ab$ is a square, forcing $c(c-1)$ to be a square. Coprimality of $c$ and $c-1$ forces in turn $c$ and $c-1$ to be both squares. This is impossible, since two squares of non-zero integers are at least 3 distance apart. Consequently, $a$ is coprime to $b$, so they are both squares.

Since $a$ and $b$ are squares, $a + 2$ and $b + 2$ are not; and since $(a + 2)(b + 2)$ is a square, $a + 2$ is not coprime to $b + 2$ (otherwise they would both be squares). Clearly, $\gcd(a+2, b+2)$ divides $(b+2)-(a+2) = b-a = q$, so $\gcd(a+2, b+2) = q$. As before, $a + 2 = (c' - 1)q$ and $b + 2 = c'q$ for some integer $c' > 1$, and we reach a contradiction along the same lines. This ends the proof.