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Write all the numbers that are perfect numbers and also a factorial

Attempt:

I know that perfect numbers have the general formula $\ \frac{2^p}{2}(2^p-1) $ where p is prime and the factorial have the general formula of $\ x!$. So the basic attempt is to find an intersection of both graphs which comes at 6 and after that both graphs diverge.

However I want to know whether there is a rigorous method to solve this question which doesnt involve graphing.

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    1. We only know that the _even_ perfect numbers are of the form you give. 2. Just because $2^{p-1}(2^p-1)$ grows slower than $x!$, it doesn't mean that they have to be different, since $p$ and $x$ are not the same. 3. Try showing that $3\mid x!$ and $3\nmid 2^{p-1}(2^p-1)$ for $p,x>2$, so they can't be equal. – 2017-01-21
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    The formula $2^{p-1}(2^p-1)$, for $p$ prime such that $2^p-1$ is itself prime, gives all *even* perfect numbers ... – 2017-01-21
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    @Wojowu I am not sure what you mean by your last sentence – 2017-01-21

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For $n \geq 4$, $n!$ is divisible by $24$, so $n/2$, $n/3$, and $n/4$ are proper positive divisors of $n$. Since $$ \frac{n}{2} + \frac{n}{3} + \frac{n}{4} = \frac{13n}{12} > n, $$ $n!$ cannot be a perfect number for $n \geq 4$. So $3! = 6$ is the only perfect number which is also a factorial.