Solution:

Let $a$ be Ian's age. Then
$$p(x) = (x - a) q(x)$$
where $q(x)$ is a polynomial with integer coefficients.

Since $p(7) = 77$, we have
$$p(7) = (7 - a) q(7) = 77 = 7 \cdot 11$$
Since $q(7)$ is an integer and $7 - a < 0$, we restrict
$$a - 7 \in \{1, 7, 11, 77\}$$
Let $b$ be the second number mentioned by Marco. Since $p(b) = 85$, we have
$$p(b) = (b - a) q(b) = 85 = 5 \cdot 17$$
Since $q(b)$ is an integer and $b - a < 0$, we restrict
$$a - b \in \{1, 5, 17, 85\}$$
Finally, we know from algebra that $b - 7$ is a divisor of $p(b) - p(7) = 85 - 77 = 8 = 2^3$. It follows that
$$b - 7 \in \{1, 2, 4, 8\}$$
Considering all the possibilities from $(\star\star)$ and $(\star\star\star)$, since $a - 7 = (a - b) + (b - 7)$, we get
$$a - 7 \in \{2, 3, 5, 6, 7, 9, 13, 18, 19, 21, 25, 86, 87, 89, 93\}$$
Recalling $(\star)$, we get $a - 7 = 7$ or $a = 14$.