Finding next perfect number - brute force
You get:


  Hint: you’ll need a sum-of-factors subprocedure.


So I would make sum-of-factors a separate procedure. It's a little confusing as is, since sumOfFactors takes a single argument - and that argument isn't the number whose factors you're summing. That is:

(define (sum-of-factors n)
    ...)

(define (next-perf n)
    (if (= (sum-of-factors n) n)
        n
        (next-perf (+ n 1)))


Note that hyphenated-name is the preferred syntax for function names, as opposed to camelCase. 

Once we're there, you can write sum-of-factors to do the looping:

(define (sum-of-factors n)
    (define (add-factor fac) (if (= (remainder n fac) 0) fac 0))
    (define (accum fac) (if (= fac n) 0 (+ (add-factor fac) (accum (+ 1 fac)))
    (accum 1))


I split it up into a few more bite-size functions. I find that helps readability. 



In Racket (which I guess isn't up for discussion), you can do the latter way better since there's an in-range procedure that gives you want you need:

(define (sum-of-factors n)
    (for/sum ([fac (in-range 1 (- n 1))])
        (if (= (remainder n fac) 0) fac 0))))


Don't think there's a Scheme equivalent unfortunately, so maybe this is more of a tease than an answer.