Q: Prove this fact: If |m| > 1, and if m has no divisor d with $\displaystyle 1 < d \leq \sqrt{|m|} $ then m must be prime.

One idea I had was:

Suppose m is not prime then m must have a divisor d with $\displaystyle 1 < d \leq \sqrt{|m|} $.

Also, I've been trying to think of a useful contradiction to use but I can't come up with any.