I add this as a new answer as it has a somewhat different flavor from my previous response. Moreover the bounds given there may still be useful. Thanks also to Edgardo for a useful discussion.
Again consider the case when $n$ is fixed and $m$ tends to infinity. To each natural number below $n$ associate a vector in ${\Bbb Z}^{\pi(n)}$ with non-negative coordinates corresponding to the exponents in the prime factorization of that number. Thus when $n=5$ we have the vectors $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(2,0,0)$ and $(0,0,1)$. Let ${\mathcal C}(n)$ denote the convex hull of these $n$ vectors. If a number $N$ is written as a product of $m$ numbers below $n$, then the prime factorization of $N$ gives a vector which is contained in the set ${\mathcal C}(n)$ dilated by $m$. Therefore it follows that the $P(m,n)$ is at most the number of lattice points contained in $m\cdot {\mathcal C}(n)$ which is asymptotically $$ m^{\pi(n)} \text{Vol}({\mathcal C}(n)). $$
Further note that any vector in $(m-n)\cdot {\mathcal C}(n)$ can be expressed as a sum of at most $m$ of the $n$ vectors corresponding to $1$ to $n$. Therefore $P(m,n)$ is at least as large as the number of lattice points contained in $(m-n) \cdot {\mathcal C}(n)$ and this also is asymptotically $$ m^{\pi(n)} \text{Vol}({\mathcal C}(n)). $$
Thus we have shown that $P(m,n) \sim m^{\pi(n)} \text{Vol}({\mathcal C}(n))$. My previous answer may be seen as giving upper and lower bounds for the volume of ${\mathcal C}(n)$ by bounding it from above and below by simplices.
It still remains interesting to ask whether the answer is always a nice polynomial in $m$. Perhaps it is the Erhart polynomial of $m\cdot {\mathcal C}(n)$? There is of course an extensive literature on this (e.g. work of Barvinok and Woods), but I'm not an expert here.