Seventy Seven Sevens

Jelly, 21 20 19 18 bytes

R7ẋḌµ;ŒċP€⁹f€FµÐLḟ

Note that the output doesn't match the OP's. I've left a comment.

Try it online!

How it works

R7ẋḌµ;ŒċP€⁹f€FµÐLḟ  Main link. Left argument: n. Right argument: l

R                   Range; yield [1, ..., n].
 7ẋ                 Times; yield [[7], ..., [7] * n].
   Ḍ                Undecimal; yield s := [7, 77, ...].
    µ         µÐL   Begin a new chain with argument s and call the chain between 
                    until the results no longer chain.
                    Return the last unique result.
      Œċ            Combinations; return all unordered pairs in integers in the
                    return value.
     ;              Concatenate the return value and its pairs.
        P€          Take the product of each individual integer and each pair in
                    the result.
          ⁹f€       Filter each; for each j in [1, ..., l], intersect [j] with the
                    array of products. The result is sorted and contains no 
                    duplicates.
                 ḟ  Filterfalse; remove the elements of s from the result.

Python 2, 116 113 109 bytes

n,l=input()
r=t={1}
exec't|={10**n/9*7};n-=n>1;r=r|{x*y for x in r for y in t if l/x/y};'*l
print sorted(r-t)

Note that TIO doesn't have enough memory for the last test case.

Try it online!

JavaScript (ES6), 103 101 bytes

Takes input in currying syntax (n)(l).

n=>l=>(a=[],g=(n,m,p,i)=>(p>l||g(n,m,(a[i>1?p:a]=p)*m,-~i),--n?g(n,m+7,p,i):a.filter(n=>n)))(n,'7',1)

Test cases

The last test case may take a few seconds to complete.

let f =

n=>l=>(a=[],g=(n,m,p,i)=>(p>l||g(n,m,(a[i>1?p:a]=p)*m,-~i),--n?g(n,m+7,p,i):a.filter(n=>n)))(n,'7',1)

console.log(f(1)(49))
console.log(f(1)(343))
console.log(f(2)(6000))
console.log(f(3)(604000))