Vyxal, 4 bytes

K⊐∑= 

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That is: does the sum (∑) of the proper divisors of the implicit input (K⊐ -> K includes the input as a divisor, so we need to remove it by returning divisors[:-1] - which is what ⊐ does) equal (=) the now cycled through implicit input (if EOI is reached, and input is taken through arguments, Vyxal resets the input pointer to 0 and continues).

Vyxal, 3+1 3 bytes

∆K= 

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-1 byte thanks to @Deadcode for telling me to not be an idiot and remember that I added a sum of proper divisors built-in. I actually forgot that I did

This is: does the sum of the proper divisors of the input (∆K) equal the input itself (=).

Vyxal, 4 bytes

K⊐∑= 

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Just the same old "does the sum of the divisors equal the input"

Vyxal, 4 bytes

K⊐∑= 

Try it Online!

That is: does the sum (∑) of the proper divisors of the implicit input (K⊐ -> K includes the input as a divisor, so we need to remove it by returning divisors[:-1] - which is what ⊐ does) equal (=) the now cycled through implicit input (if EOI is reached, and input is taken through arguments, Vyxal resets the input pointer to 0 and continues).