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.

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.

Edit: Here is a better upper bound for $P(m,n)$: $$ P(m,n) \le (1+o(1)) \frac{m^{\pi(n)}}{\pi(n)!} \prod_{p\le n} \lfloor \frac{\log n}{\log p}\rfloor (\pi(n))^{\pi(\sqrt{n})}. $$ Combining the upper and lower bounds, we now have for fixed $n$ and $m$ large $$ P(m,n) = \frac{m^{\pi(n)}}{\pi(n)!} \exp(O(\sqrt{n})). $$

Now for the proofs. For the original 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.

Now for the improved upper bound. Divide the primes below $n$ into the sets $P_{k}=\{n^{1/(k+1)}<p\le n^{1/k}\}$ for $1\le k\le \log n/\log 2$. Suppose a number $N=\prod_{p\le n} p^{a_p}$ is counted in $P(m,n)$. Then we see that for each $1\le k\le \log n/\log 2$ we must have $\sum_{p\in P_k} a_p \le km$. The number of non-negative $a_p$ with $\sum_{p\in P_k} a_p\le km$ is $\binom{km+|P_k|}{km}=\binom{km+|P_k|}{|P_k|}$. Thus we conclude that $$ P(m,n) \le \prod_{k\le \log n/\log 2} \binom{km+|P_k|}{|P_k|}. $$ Now for $k=1$ we have $$ \binom{m+|P_1|}{|P_1|} = (1+o(1)) \frac{m^{|P_1|}}{|P_1|!} \le (1+o(1)) \frac{m^{|P_1|}}{\pi(n)!} \pi(n)^{\pi(\sqrt{n})}. $$ And for larger $k$ simply use that $$ \binom{km+|P_k|}{|P_k|} \le (1+o(1))(km)^{|P_k|}. $$ The new upper bound follows.