Smallest multiplier that reveals a factor of a semiprime

05AB1E, 18 16 15 bytes

-2 bytes thanks to Riley!

-1 byte thanks to Emigna!

[N¹*b¹Ñ¦¨båOiNq

Explanation:

[                   # Infinite loop start
 N                  # Push the amount of times we have iterated
  ¹*               # Multiplied by input
    b              # Convert to binary
     ¹Ñ¦¨b         # Calculate the proper divisors of the input in binary excluding one
          åO       # Check if a substring of N * m in binary is in the divisors
            iNq    # If so, print how many times we have iterated and terminate the program

Try it online!

Pyth, 13 bytes

ff}.BY.B*TQPQ

Demonstration

Explanation:

ff}.BY.B*TQPQ
f                Find the first integer >= to 1 where the following is true
 f         PQ    Filter the prime factors of the input
        *TQ      Multiply the input by the outer integer
      .B         Convert to a binary string
   .BY           Convert the prime factor to a binary string
  }              Check whether the factor string is in the multiple string.

JavaScript (ES6), 96 95 80 bytes

n=>F=m=>(k=p=>p&&(q=1,g=x=>1<x&&x<n&n%x<1|g(x>>1,q*=2))(p)|k(p-q))(n*m)?m:F(-~m)

A function which returns a recursive function which uses a recursive function which uses a recursive function. I'm really starting to wonder if the .toString(2) route would be shorter...

Assign to a variable e.g. f=n=>... and call with an extra pair of parens, f(9)(). If that's not allowed (the meta post is at +6/-2), you can use this 83-byte version with standard invocation:

f=(n,m)=>(k=p=>p&&(q=1,g=x=>1<x&&x<n&n%x<1|g(x>>1,q*=2))(p)|k(p-q))(n*m)?m:f(n,-~m)

Both versions work for all but the last three test cases. You can try these test cases as well by changing x>>1 to (x-x%2)/2.