smc

Since there are only $3$ candidate values for $m$, we test $0,12,-12$. If $m=0$, then the expression is $x^2+3xy+x$. The term $x$ appears in each monomial, giving $x(x+3y+1)$, which has two factors that are linear in $x$ and $y$ with integer coefficients. This eliminates $C$ and $D$. We next test $m=12$. This gives $x^2+3xy+x+12y-12$. The linear factors will be of the form $(Ax+By+C)(Dx+Ey+F)$. We will match coefficients from the skeleton form to the expression that is to be factored. Matching the $y^2$ coefficient gives $BE=0$, and WLOG we can let $B=0$ to give $(Ax+C)(Dx+Ey+F)$. Matching the $x^2$ coefficients gives $AD=1$. Since $A,D$ are integers, we either have $(A,D)=(1,1),(-1,-1)$. WLOG we can pick the former, since if $(a)(b)$ is a factorization, so is $(-a)(-b)$. We now have $(x+C)(x+Ey+F)$. Matching the $xy$ term gives $E=3$. We now have $(x+C)(x+3y+F)$. Matching the $y$ term gives $3C=12$, which leads to $C=4$. Finally, matching constants leads to $4F=-12$ and $F=-3$ for a provisional factorization of $(x+4)(x+3y-3)$. We've matched every coefficient except for $x$, and finding the $x$ term by selectively multiplying leads to $4x+-3x=x$, which matches, so this is a real factorization. We now must check $m=-12$. The expression is $x^2+3xy+x-12y+12m$, and the skeleton once again is $(Ax+By+C)(Dx+Ey+F)$. The beginning three steps matching $y^2,x^2,xy$ are the same, leading to $(x+C)(x+3y+F)$. Matching the $y$ term gives $3C=-12$, or $C=-4$. The skeleton becomes $(x-4)(x+3y+F)$. The constant locks in $F=-3$ for a provisional factorization of $(x-4)(x+3y-3)$. However, the $x$ term does not match up, because it is $-4x+-3x=-7x$, when it needs to be $x$. Thus, $m=0,12$ work and $m=-12$ does not, giving answer $\boxed{0,12}$.