Pretty Smooth Moves

Jelly, 12 bytes

Æf>½S
³²ḊÇÐḟ

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How it works

³²ḊÇÐḟ  Main link. No arguments.

³       Yield 100.
 ²      Square it to yield 10,000.
  Ḋ     Dequeue; yield [2, ..., 10,000].
   ÇÐḟ  Filter-false; keep elements for which the helper link returns 0.

Æf>½S   Helper link. Argument: n

Æf      Compute the prime factorization of n.
  >½    Compare the prime factors with the square root of n.
    S   Sum; add the resulting Booleans.

Brachylog, 21 19 bytes

1 byte thanks to Fatalize, for inspiration of another 1 byte.

100^:4reP$ph^<=P@w\

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Takes about 6 seconds here.

100^:4reP$ph^<=P@w\
100                      100
   ^                     squared
    :4                   [10000,4]
      r                  [4,10000]
       eP                P is an integer in that interval (choice point),
        P$ph^<=P         P, prime factorized (from biggest to smallest),
                         take the first element, squared, is less than
                         or equal to P
               P@w       Write P with a newline,
                  \      Backtrack to the last choice point and make
                         a different choice until there is no more
                         choice and the program halts.

Previous 21-byte solution

100^:4reP'($pe^>P)@w\

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Takes about 6 seconds here.

100^:4reP'($pe^>P)@w\
100                      100
   ^                     squared
    :4                   [10000,4]
      r                  [4,10000]
       eP                P is an integer in that interval (choice point),
        P'(      )       The following about P cannot be proved:
           $pe               one of its prime factor
              ^              squared
               >P            is greater than P
                  @w     Write P with a newline,
                    \    Backtrack to the last choice point and make
                         a different choice until there is no more
                         choice and the program halts.

Haskell, 53 bytes

r=[1..10^4]
[n|n<-r,product[x|x<-r,x*x<=n]^n`mod`n<1]

I don't have time to golf this now, but I want to illustrate a method for testing if n is pretty smooth: Multiply the numbers from 1 to sqrt(n) (i.e. compute a factorial), raise the product to a high power, and check if the result is a multiple of n.

Change to r=[2..10^4] if 1 should not be output.