How many ways to write N as a product of M integers?

Python 3, 59

f=lambda N,M,i=2:i<=N and f(N/i,M-1,i)+f(N,M,i+1)or-~M==N<2

We count up potential divisors i. With the additional argument i as the lowest allowed divisor, the core recursive relation is

f(N,M,i)=f(N/i,M-1,i)+f(N,M,i+1)

For each i, we either choose to include it (possible as a repeat), in which case we divide the required product N by i and decrement M. If we don't, we increase i by 1, but only if i<N, since there's no use checking divisors greater than N.

When the minimum divisor i exceeds N, there's no more potential divisors. So, we check if we've succeeded by seeing if M==0 and N==1, or, equivalently, M+1==N==1 or M+1==N<2, since when M+1==N, the mutual value is guaranteed to be a positive integer (thanks to FryAmTheEggman for this optimization).

This code will cause a stack overflow for N about 1000 on most systems, but you can run it in Stackless Python to avoid this. Moreover, it is extremely slow because of its exponential recursive branching.

Pyth - 24 23 22 21 bytes

Not a complicated solution. Will be golfing more. Just takes cartesian product of lists and filters. Same strategy as @optimizer (I'm guessing because of his comment, didn't actually decipher that CJam) Thanks to @FryAmTheEggman for 2 bytes and trick with M.

Ml{m`Sdfqu*GHT1G^r2GH

Defines a function g with args G and H

M                    function definition of g with args G and H
 l                   length of
  {                  set (eliminates duplicates)
   m                 map
    `Sd              repr of sorted factors so can run set (on bash escape ` as \`)
    f                filter
     q      G        equals first arg
      u*GHT1         reduce by multiplication
     ^     H         cartesian product by repeat second arg
       r2G           range 2 to first arg

Worked on all test args except last, was too slow on that one but there is no time limit given.

Ruby, 67

f=->n,m,s=2,r=0{m<2?1:s.upto(n**0.5){|d|n%d<1&&r+=f[n/d,m-1,d]}&&r}

Actually reasonably efficient for a recursive definition. For each divisor pair [d,q] of n, with d being the smaller one, we sum the result of f[q,m-1]. The tricky part is that in the inner calls, we limit factors to ones greater than or equal to d so that we don't end up double-counting.

1.9.3-p327 :002 > f[30,2]
 => 3 
1.9.3-p327 :003 > f[2310,4]
 => 10 
1.9.3-p327 :004 > f[15,4]
 => 0 
1.9.3-p327 :005 > f[9,2]
 => 1