= Solution
Use the dense standard <affine charts>. Their product is
$$
\mathbb A^n\times\mathbb A^m\cong\mathbb A^{n+m},
$$
which is also a dense <affine chart> of $\mathbb P^{n+m}$. The identification of these charts gives \b[a birational isomorphism] between the two <projective varieties>.
To distinguish them, calculate their <divisor class groups> explicitly. Let $H\subset\mathbb P^r$ be a coordinate <hyperplane>. Its complement is $\mathbb A^r$, whose <coordinate ring> is a <unique factorization domain>. Any <Weil divisor> on $\mathbb P^r$ restricts on this chart to a <principal Weil divisor>; subtract that <principal Weil divisor> on $\mathbb P^r$, leaving a <Weil divisor> supported on $H$. Thus $[H]$ generates $\operatorname{Cl}(\mathbb P^r)$. If $aH=\operatorname{div}(q)$, then $q$ has zero <Weil divisor> on $\mathbb A^r$. A reduced fraction in a <unique factorization domain> with zero orders at every <prime Weil divisor> must be a <unit>, so $q\in k^*$. It follows that $a=0$. We have proved
$$
\boxed{\operatorname{Cl}(\mathbb P^r)\cong\mathbb Z\quad(r>0).}
$$
In the product, let $H_1$ and $H_2$ be coordinate <hyperplanes> in the factors and put
$$
D_1=H_1\times\mathbb P^m,\qquad D_2=\mathbb P^n\times H_2.
$$
The complement of $D_1\cup D_2$ is $\mathbb A^{n+m}$, again with a <unique factorization domain> as its <coordinate ring>. The same restriction-and-subtraction argument says that $[D_1],[D_2]$ generate the <divisor class group>. If
$$
aD_1+bD_2=\operatorname{div}(q),
$$
then on that <affine chart> $q$ has neither zeros nor poles along any <prime Weil divisor>. <Unique factorization> makes $q$ a <unit> of $k[s_1,\ldots,s_n,t_1,\ldots,t_m]$, hence a constant. It follows that $a=b=0$. Therefore
$$
\boxed{\operatorname{Cl}(\mathbb P^n\times\mathbb P^m)\cong\mathbb Z^2.}
$$
Both spaces are <smooth varieties>, so their <divisor class groups> also equal their <Picard groups>. An <isomorphism> preserves the <divisor class group>; $\mathbb Z^2$ and $\mathbb Z$ are not <isomorphic>. Thus \b[the spaces are birational but not isomorphic]. The assumptions $n,m>0$ ensure that both boundary <prime Weil divisors> actually occur.
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