Something odd with this formulation. The version I remember was
$$
107^{56}(m^2-1)+2m+3 = \binom{113^{114}}{n}.
$$I will assume this is the correct one, and will provide a proof. Let $n\in\{1,2,\dots,113^{114}-1\}$. Then, $113\mid \binom{113^{114}}{n}$. Hence,
$113\mid 107^{56}(m^2-1)-2m+5$. Now, $107^{56}\equiv 6^{56}\pmod{113}$. Note that by Fermat's theorem, $6^{56}\in\{-1,1\}\pmod{112}$, with being one iff $6$ is a quadratic residue modulo $113$, and $-1$ otherwise. I now show $6$ is not a quadratic residue modulo $113$. Observe that, $113\equiv 1\pmod{8}$, and $113\equiv -1\pmod{6}$. Thus, $2$ and $-1$ are both quadratic residues, whereas $-3$ is not. Thus, $6$ is not, either. Therefore, $6^{56}\equiv -1\pmod{113}$. With this, we conclude that $-m^2+1+2m+3\equiv 0\pmod{113}\iff (m-1)^2\equiv 5\pmod{113}$. Now using quadratic reciprocity, it follows that $5$ is not a quadratic residue modulo $113$. Hence, this is impossible.
The remaining possibilities are $n\in\{0,113^{114}\}$. Both cases are easily studied to show to yield no solutions.