Assume that $2560m^{2} +5m+6=n^{5} \cdots ( \heartsuit )$ for some $( n,m) \in \mathbb{Z}^{2}$. First, by taking $( \heartsuit )$ modulo $5$, we have $1\equiv n^{5} \equiv n$. Let $n=5k+1$ for some $k\in \mathbb{Z}$. Note that $n^{5} =( 5k+1)^{5} \equiv 1\pmod{25}$. Second, by taking $( \heartsuit )$ modulo $25$, we have $10m^{2} +5m+6\equiv 1$. So we must have $5\left( 2m^{2} +m+1\right) \equiv 0$. It follows that $5\mid 8\left( 2m^{2} +m+1\right)$, which implies $5\mid ( 4m+1)^{2} +7$. This is impossible because $-2$ is not a quadratic residue modulo $5$. (Note that $1^{2} \equiv 1,2^{2} \equiv -1,3^{2} \equiv -1,4^{2} \equiv 1\pmod 5$)$ $

$ mod 6 $