Hi.

Let us work with natural numbers (so |m|=m) because e.g. -3 satisfies the assumption but is not a prime.

Suppose that m>1 has no proper divisor d such that 1<d<=sqrt(m), and that m is not prime.

Then there is some divisor k of m, k distinct from m and 1.

So m=k*j, where j is also distinct from m and 1.

By hypothesis we have k>sqrt(m), j>sqrt(m).

Multiplying these inequalities we get k*j > sqrt(m)*sqrt(m) = m, which contradicts m=k*j.