It is not very hard to see that for each prime power $q$ and natural numbers $n,h$, we have an embedding
$$\iota \colon \mathrm{GL}(n,q^h) \hookrightarrow \mathrm{GL}(nh, q),$$
obtained by choosing a basis for the finite field $\mathbb{F}_{q^h}$ over the base field $\mathbb{F}_q$.
It is not very hard either to see that this does *not* induce an embedding of the corresponding projective groups $\mathrm{PGL}(n,q^h)$ into $\mathrm{PGL}(nh, q)$.
Of course, this does not exclude the possibility that $\mathrm{PGL}(n,q^h)$ is nevertheless isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$, and that is precisely my question:

For which values of $q,n,h$ with $h > 1$ is $\mathrm{PGL}(n,q^h)$ isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$?

Notice that comparing the order of both groups does not rule out anything at all.

I believe I read somewhere that $\mathrm{PGL}(n,q^h)$ is isomorphic to a subgroup of $\mathrm{PGL}(n^h, q)$ (but I forgot the precise construction), so this would at least tell that the answer is positive for $n=h=2$. (**Edit**: This fact is now confirmed by Derek Holt's answer.)