There exists a unique isomorphism $M \otimes N \to N \otimes M$
It is not true that $n\otimes m = 0$ implies either $n$ or $m =0$ (see example below). To prove injectivity you should define a map going the other way and show that these maps are inverse.
example:
$\bar1\otimes \bar2 \in \mathbb{Z}/2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}/3\mathbb{Z}$ satisfies $\bar1\otimes \bar2=\bar1\otimes (2\cdot\bar1)=(\bar1\cdot 2)\otimes \bar1= \bar0\otimes \bar1=0$ but $\bar1\in\mathbb{Z}/2\mathbb{Z}$ and $\bar2\in\mathbb{Z}/3\mathbb{Z}$ are not zero.