The answer is $759$.

For $m = 1, 2, 3$, the left-hand side becomes $\left[\frac{3^n}{4}\right]$, $\left[\frac{5^n}{14}\right]$, $\left[\frac{7^n}{30}\right]$. Since
$$\left[\frac{3^8}{4}\right] < 2013 < \left[\frac{3^9}{4}\right], \quad \left[\frac{5^6}{14}\right] < 2013 < \left[\frac{5^7}{14}\right], \quad \left[\frac{7^5}{30}\right] < 2013 < \left[\frac{7^6}{30}\right],$$
there is no solution.

For $m \ge 4$, we have
$$2^{11} > 2014 > \frac{(2m+1)^n}{m(3m+1)} > \frac{(2m)^n}{m(4m)} = 2^{n-2}m^{n-2} \ge 2^{n-2}4^{n-2} = 2^{3n-6}.$$
This implies $n \le 5$.

* For $n \le 2$, we have $\frac{(2m+1)^n}{m(3m+1)} < \frac{(3m)^2}{m(3m)} = 3 < 2013$.

* For $n = 3$, we have
$$\frac{(2m+1)^3}{m(3m+1)} = \frac{8m^3 + 12m^2 + 6m + 1}{3m^2 + m} = \frac{8}{3}m + \frac{28}{9} + \frac{26m+9}{9(3m^2 + m)}.$$
Clearly, the last fraction is less than $1$. So we need $2012 < \frac{8}{3}m + \frac{28}{9} < 2014$.
The only integer solution is $m = 754$. We check that
$$\begin{aligned} & \left[ \frac{8}{3}(754) + \frac{28}{9} + \frac{26(754) + 9}{9(3(754)^2 + (754))} \right] \\ &= \left[ 2013 + \frac{7}{9} + \frac{26(754) + 9}{9(3(754)^2 + (754))} \right] \\ &= 2013 \end{aligned}$$
since $\frac{26(754) + 9}{9(3(754)^2 + (754))} < \frac{27(754)}{27(754)^2} < \frac{2}{9}$. So $m = 754$ is a solution.

* For $n \ge 4$, we claim that $f(m) = \frac{(2m+1)^n}{m(3m+1)}$ is increasing in $m$. Note that
$$\ln f(m) = n \ln(2m+1) - \ln m - \ln(3m+1),$$
and so
$$\frac{f'(m)}{f(m)} = \frac{2n}{2m+1} - \frac{1}{m} - \frac{3}{3m+1} \ge \frac{8}{2m+1} - \frac{3}{2m+1} - \frac{3}{2m+1} > 0.$$
Thus, $f'(m) > 0$, so that $f$ is increasing.

Now, for $n = 4$, we check that
$$\frac{(2(18) + 1)^4}{18(3(18) + 1)} = \frac{37^4}{18 \cdot 55} = \frac{1369^2}{990} < 2013$$
and
$$\frac{(2(19) + 1)^4}{19(3(19) + 1)} = \frac{39^4}{19 \cdot 58} = \frac{1521^2}{1102} > 2014.$$
There is no solution.

$$\frac{(2(4) + 1)^5}{4(3(4) + 1)} = \frac{9^5}{4 \cdot 13} = \frac{59049}{52} < 2013,$$
$$\left[ \frac{(2(5) + 1)^5}{5(3(5) + 1)} \right] = \left[ \frac{11^5}{5 \cdot 16} \right] = \left[ \frac{161051}{80} \right] = 2013$$
and
$$\frac{(2(6) + 1)^5}{6(3(6) + 1)} = \frac{13^5}{6 \cdot 19} = \frac{371293}{114} > 2014.$$
Thus, $m = 5$ is another solution.

To summarize, the answer is $754 + 5 = 759$.