The solutions are $(m, n) = (1, 0), (1, 113^{114})$.
When $n = 0$ or $n = 113^{114}$, the equation becomes $107^{56}(m^2 - 1) - 2m + 3 = 1$. This implies
$$(m - 1)(107^{56}(m + 1) - 2) = 0.$$
Clearly, $107^{56}(m + 1) - 2 = 0$ has no solution. Thus, the only solution is $m = 1$.

When $0 < n < 113^{114}$, by Lucas' theorem, $113$ divides $\binom{113^{114}}{n}$. This also holds when $n > 113^{114}$ since the binomial coefficient is 0. Therefore, we have
$$107^{56}(m^2 - 1) - 2m + 3 \equiv 0 \pmod{113}. \qquad (1)$$
Let $g$ be a primitive root modulo $113$, and let $107 \equiv g^a \pmod{113}$. Using the Legendre symbol and the quadratic reciprocity law, we find that
$$\left(\frac{107}{113}\right) = \left(\frac{-6}{113}\right) = \left(\frac{-1}{113}\right)\left(\frac{2}{113}\right)\left(\frac{3}{113}\right) = (1)(1)\left(\frac{113}{3}\right) = \left(\frac{2}{3}\right) = -1.$$
This shows $107$ is a quadratic nonresidue modulo $113$, and so $a$ is odd. Now, note that
$$107^{56} \equiv g^{56a} \equiv -1 \pmod{113}$$
since $(g^{56a})^2 = g^{112a} = g^{\varphi(113)a} \equiv 1 \pmod{113}$ and $112 + 56a$. Therefore, (1) becomes
$$-(m^2 - 1) - 2m + 3 \equiv 0 \pmod{113}.$$

This is the same as $(m+1)^2 \equiv 5 \pmod{113}$. However, we check that
$$\left(\frac{5}{113}\right) = \left(\frac{113}{5}\right) = \left(\frac{3}{5}\right) = -1,$$
which implies 5 is a quadratic nonresidue modulo 113. This is a contradiction. Therefore, there is no solution.