We must have $m^{2}+6 m+28=n^{2}$, where $n$ is an integer. Rewrite this as $(m+3)^{2}+19=$ $n^{2} \Rightarrow n^{2}-(m+3)^{2}=19 \Rightarrow(n-m-3)(n+m+3)=19$. Let $a=n-m-3$ and $b=n+m+3$, so we want $a b=19$. This leaves only 4 cases: - $a=1, b=19$. Solve the system $n-m-3=1$ and $n+m+3=19$ to get $n=10$ and $m=6$, giving one possible solution. - $a=19, b=1$. Solve the system, as above, to get $n=10$ and $m=-12$. - $a=-1, b=-19$. We get $n=-10$ and $m=-12$. - $a=-19, b=-1$. We get $n=-10$ and $m=6$. Thus the only $m$ are 6 and -12 .