I'll consider the problem when $n$ is fixed and $m$ tends to infinity.  Here the examples provided by the OP seem to suggest that $P(m,n)$ behaves like a polynomial in $m$ of degree $\pi(n)$.  I'll prove that (again thinking that $n$ is small, and $m$ is large) 
$$ 
(1+o(1)) \frac{m^{\pi(n)}}{\pi(n)!} \prod_{p\le n} \lfloor \frac{\log n}{\log p}\rfloor 
\le P(m,n) \le (1+o(1)) \frac{m^{\pi(n)}}{\pi(n)!} \prod_{p\le n} \frac{\log n}{\log p}.
$$ 
For example when $n=10$ this shows that 
$$ 
P(m,10) \le (1+o(1)) \frac{m^4}{24} \frac{(\log 10)^4}{(\log 2)(\log 3)(\log 5)(\log 7)} = (0.4911\ldots+o(1)) m^4,
$$ 
and 
$$ 
P(m,10) \ge (1+o(1)) \frac{m^4}{4},
$$ 
which are in keeping with the asymptotic $P(m,10) \sim m^4/3$ found in the question.  As $n$ gets larger (but still in the range where $m$ is much larger than $n$) the upper and lower bounds given above differ at most by a factor of $2^{\pi(n)}$ (and using the prime number theorem they really differ by a factor of $e^{(1/2+o(1))n/\log n}$).  It may be interesting to obtain more refined asymptotics here.  

Now for the proofs.  For the upper bound, as observed by Eric Naslund, one has $P(m,n) \le \Psi(n^m,n)$ where $\Psi(x,y)$ denotes the number of integers below $x$ all of whose prime factors lie below $y$.  Naturally $\Psi(x,y)$ has been extensively studied.  In the range we are interested in, $y$ is kept very small (less than $\log x$) and $x$ is much bigger.  Here the problem of counting smooth numbers should be thought of as a problem of counting lattice points inside a high dimensional ``tetrahedron."  This is an old idea, and for smooth numbers was worked out by Ennola.  There is a paper by Granville in Aequationes Math. which discusses such lattice point problems in many related situations: see http://www.dms.umontreal.ca/~andrew/PDF/tetrahedron.pdf (specifically, eqn (2.4) there).  Note that $\Psi(n^m,n)$ counts all lattice points in $\pi(n)$ dimensional space such that the coordinates are nonnegative and $\sum_{p\le n} a_p \log p \le m\log n$.  The number of such lattice points is asymptotically the volume of this tetrahedron, and this gives our upper bound.  

As for the lower bound, I claim that any number of the form $\prod_{p\le n} p^{a_p}$ with the property that $a_p$'s are non-negative integers and  
$$   
\sum_{p\le n} a_p \Big(\lfloor \frac{\log n}{\log p} \rfloor\Big)^{-1} \le m-\pi(n),
$$ 
is counted in $P(m,n)$.  To see this, for each prime $p\le n$ put $b_p=\lfloor \log n/\log p\rfloor$ and use $[a_p/b_p]$ values of $p^{b_p}$ and one extra power of $p$ to get the exponents to add up to $a_p$; the assumed inequality guarantees that at most $m$ numbers are used in doing this. So once again we are counting the lattice points in a tetrahedron, and computing the volume of this tetrahedron we obtain the lower bound.