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Suppose  
\[n\]
  is square free.  
\[n\]
  is not divisible by any square number or the square of any prime number.
\[n\]
  must be of the form  
\[n=p_1p_2...p_k\]
  then using Euler's Totient Function the number of divisors is  
\[\tau (n)=\underbrace{(1+1)...(1+1)}_{k \; times}=2^k\]
.
The converse is not true. That is, if  
\[\tau (n)=2^m\]
  then  
\[n\]
  need not be square free. For example,  
\[\tau (135)=\tau (3^3 \times 5)(1+3)(1+1)=2^3\]
  but 135 is divisible by 9.