I'm afraid I'm getting spoiled by my CAS and got the answer initially just by writing a loop. But reflecting a bit and using reference
Divisor Function -- from Wolfram MathWorld
note that s(ab) = s(a)s(b) and that s(p) = p+1 where p prime. So s(p_1*p_2*...*p_i) = (p_1+1)(p_2+1)...(p_i+1). So just build the sequence and stop when the condition is met.
The answer to the second question is yes, there are many such integers m.