Why does bromide not react with water when methyl ammonium bromide is dissolved in water?
The key is comparing $\mathrm{p}K_\mathrm{a}$ values. A good resource for this is the Evans $\mathrm{p}K_\mathrm{a}$ table. Unfortunately, the table does not contain $\ce{CH3NH3+}$, but in the section protonated nitrogen there is a value quoted for $\ce{C2H5NH3+}$ — which is chemically similar enough to methyl ammonium to assume the $\mathrm{p}K_\mathrm{a}$ values to be essentially identical.

\begin{array}{lr}\hline
\text{compound} &amp; \mathrm{p}K_\mathrm{a}\\ \hline
\ce{EtNH3+} &amp; 10.6\phantom{0}\\
\ce{HBr} &amp; -9.00 \\
\ce{H3O+} &amp; -1.7\phantom{0} \\ \hline\end{array}

Notice how the $\mathrm{p}K_\mathrm{a}$ values differ by almost $20$ logarithmic units. $\ce{HBr}$’s $\mathrm{p}K_\mathrm{a}$ value is even lower than water’s by $7$ to $9$ logarithmic units depending on the source used. Thus, while the amine exhibits extensive acid/base chemistry in water, bromide will not be protonated to any noticeable extent. Or you could say that your final equation will be strongly favouring the reactant side.

I’ll also point you to a recent answer of mine wherein I calculated the concentration of $\ce{HF}\ (\mathrm{p}K_\mathrm{a} \approx 3.17$, so notably weaker than $\ce{HBr}$) in a 1 molar solution of $\ce{NaF}$. The case is very comparable to your’s.