It looks to me like in the case where both the cardinalites are finite, there exists one solution for the first equation and one solution for the second as long as $m>=n$. Otherwise there's none.
In the case where one of them is infinte, it has to be $m$ or else both of the equations don't have a solution. In that case, it seems to me there are infinite solutions for both of the equations.
Lastly, in the case where both of them are infinite, again $m>=n$ or there are no solutions for either of the equations. Again, in this case there seem to be an infinite amount of solutions for both of the equations.
Have I made any mistakes here? How do I formally write these thoughts? Also, is there an easier way to solve problems like this?
Thanks a lot for your time!
$n\cdot x=m$ can have no solutions, even if $n\leq m$. For example $n=3$ and $m=5$. But you are right that there is at most one solution if $m$ is finite.
If $m<n$ then there are never solutions to both equations, but $n\cdot x=0$ still has one solution. So you should have separated these cases ($m=0$ and $m<n$) from the beginning, thus assuming in all the remaining cases that $n\leq m$. If $n=0$ then one equation has no solution except for $m=0$, and similarly for the case of $n=1$, so these should be handled separately as well. Now we can assume $1<n\leq m$. There is the case where $m$ is finite and another where it is infinite.
Formally speaking, my recommendation is to keep the separation for cases. If there is no solution, prove that (e.g. $n=0$ and $m=\aleph_0$ have no solutions for $n\cdot x=m$). If there is one solution, prove that it exists and it is unique. If you want to prove there are infinitely many solutions, just show that there is an infinite family of solutions (e.g. $\{n\in\omega\mid n>1\}$ usually works).