(Note: The following proof was copied verbatim from the answer of MSE user Alex Francisco.)
First, note that for any coprime $a, b \in \mathbb{N}_+$, there is$$
I(ab) = I(a) I(b).
$$
Suppose there is an even number $n = 2^k \cdot l$, where $k \geq 1$ and $l$ odd, such that$$
I(n) = \frac{p + 2}{p}.
$$
Case 1: $k \geq 2$. Then$$
\frac{p + 2}{p} = I(2^k \cdot l) = I(2^k) I(l) \geq I(2^k) = \frac{2^{k + 1} - 1}{2^k} = 2 - \frac{1}{2^k} \geq \frac{7}{4},
$$
which implies $4(p + 2) \geq 7p$, contradictory to $p \geq 3$.
Case 2: $k = 1$ and $p \geq 5$. Then, analogously,$$
\frac{p + 2}{p} \geq 2 - \frac{1}{2^k} = \frac{3}{2},
$$
which implies $2(p + 2) \geq 3p$, contradictory to $p \geq 5$.
Case 3: $k = 1$ and $p = 3$. Then$$
\frac{5}{3} = I(2l) = \frac{3}{2} I(l) \Longrightarrow I(l) = \frac{10}{9}.
$$
If $l$ has an odd prime factor $q < 10$, suppose $q^m \mathbin{\|} l$, then$$
\frac{10}{9} = I(l) = I(q^m) I\left( \frac{l}{q^m} \right) \geq I(q^m) = \frac{q^{m + 1} - 1}{q^m (q - 1)}\\
= 1 + \frac{1}{q - 1}\frac{q^m - 1}{q^m} \geq 1 + \frac{1}{q - 1}\frac{q - 1}{q} = 1 + \frac{1}{q} > \frac{10}{9},
$$
a contradiction. Now suppose the prime factorization of $l$ is$$
l = \prod_{i = 1}^s p_i^{a_i},
$$
then $p_i \geq 11$. Because$$
\frac{10}{9} = I(l) = \prod_{i = 1}^s \frac{1 + p_i + \cdots + p_i^{a_i}}{p_i^{a_i}},
$$
then$$
9 \prod_{i = 1}^s \sum_{j = 0}^{a_i} p_i^j = 10 \prod_{i = 1}^s p_i^{a_i} \Longrightarrow \left. 9 \,\middle|\, \prod_{i = 1}^s p_i^{a_i} \right.,
$$
contradictory to $p_i \geq 11$.
Therefore, there does not exist an even positive integer $n$ such that$$
I(n) = \frac{p + 2}{p}.
$$
I(ab) = I(a) I(b).
$$
Suppose there is an even number $n = 2^k \cdot l$, where $k \geq 1$ and $l$ odd, such that$$
I(n) = \frac{p + 2}{p}.
$$
Case 1: $k \geq 2$. Then$$
\frac{p + 2}{p} = I(2^k \cdot l) = I(2^k) I(l) \geq I(2^k) = \frac{2^{k + 1} - 1}{2^k} = 2 - \frac{1}{2^k} \geq \frac{7}{4},
$$
which implies $4(p + 2) \geq 7p$, contradictory to $p \geq 3$.
Case 2: $k = 1$ and $p \geq 5$. Then, analogously,$$
\frac{p + 2}{p} \geq 2 - \frac{1}{2^k} = \frac{3}{2},
$$
which implies $2(p + 2) \geq 3p$, contradictory to $p \geq 5$.
Case 3: $k = 1$ and $p = 3$. Then$$
\frac{5}{3} = I(2l) = \frac{3}{2} I(l) \Longrightarrow I(l) = \frac{10}{9}.
$$
If $l$ has an odd prime factor $q < 10$, suppose $q^m \mathbin{\|} l$, then$$
\frac{10}{9} = I(l) = I(q^m) I\left( \frac{l}{q^m} \right) \geq I(q^m) = \frac{q^{m + 1} - 1}{q^m (q - 1)}\\
= 1 + \frac{1}{q - 1}\frac{q^m - 1}{q^m} \geq 1 + \frac{1}{q - 1}\frac{q - 1}{q} = 1 + \frac{1}{q} > \frac{10}{9},
$$
a contradiction. Now suppose the prime factorization of $l$ is$$
l = \prod_{i = 1}^s p_i^{a_i},
$$
then $p_i \geq 11$. Because$$
\frac{10}{9} = I(l) = \prod_{i = 1}^s \frac{1 + p_i + \cdots + p_i^{a_i}}{p_i^{a_i}},
$$
then$$
9 \prod_{i = 1}^s \sum_{j = 0}^{a_i} p_i^j = 10 \prod_{i = 1}^s p_i^{a_i} \Longrightarrow \left. 9 \,\middle|\, \prod_{i = 1}^s p_i^{a_i} \right.,
$$
contradictory to $p_i \geq 11$.
Therefore, there does not exist an even positive integer $n$ such that$$
I(n) = \frac{p + 2}{p}.
$$
