Can the following argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?

(The following post is extracted verbatim from this MSE question.)

This is related to this earlier MSE question.

Let $\sigma(x)$ be the sum of the divisors of $x$, and denote the abundancy index of $x$ by $I(x):=\sigma(x)/x$.

If the equation $I(a) = b/c$ has no solution $a \in \mathbb{N}$, then the rational number $b/c$ is said to be an abundancy outlaw.

In the earlier question, it is shown that:

(a) If $I(n) = (p+2)/p$, then $n$ is an odd square.

(b) If $I(n) = (p+2)/p$, then $p \mid n$.

So now suppose to the contrary that $I(n) = (p+2)/p$.  From (a) and (b), since $p$ is a prime number, then $p^2 \mid n$.  It follows that

$$\frac{p^2 + p + 1}{p^2} = I(p^2) \leq I(n) = \frac{p+2}{p},$$

whence there is still no contradiction.

From the divisibility constraint $p^2 \mid n$, we have that $p^2 \leq n$.  We claim that $p^2 \neq n$.  Suppose to the contrary that $p^2 = n$.

Then we have

$$\frac{\sigma(n)}{n} = \frac{p+2}{p} = \frac{p(p+2)}{p^2}$$

so that

$$p^2 + p + 1 = \sigma(p^2) = \sigma(n) = p^2 + 2p,$$

which contradicts the fact that $p$ is an odd prime.

This implies that $p < \sqrt{n}$, which further means that

$$\gcd(n, \sigma(n)) = \frac{n}{p} > \sqrt{n} > {10}^8$$

where the lower bound $n > {10}^{16}$ is due to Richard F. Ryan, "Results concerning uniqueness for $\sigma(x)/x = \sigma(p^n q^m)/(p^n q^m)$ and related topics, International Math. J., 2002 , V2#5pp497-514".

Here is my question:

Can the preceding argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?