Suppose that $12^{x} +13^{y} -14^{z} =2013^{t}$ for $( x,y,z,t) \in \mathbb{N}^{4}$. For any $m\in \mathbb{N}$, let $E_{m}$ be the equation $12^{x} +13^{y} -14^{z} \equiv 2013^{t}\pmod m$. From $E_{3}$ , we have $1-2^{z} \equiv 0\pmod 3$, which implies $z\equiv 0\pmod 2$. From $E_{13}$ , we have $( -1)^{x} -1\equiv ( -2)^{t}\pmod{13}$, which implise $x\equiv 1\pmod 2$ and $t\equiv 1\pmod{12}$. First, we will show that $x=1$. Assume,for the sake of contradiction, that $x\geq 2$. If $z\geq 4$, then from $E_{16}$ and $t\equiv 1\pmod 4$, we have $13^{y} \equiv 13\pmod{16}$, which implise $y\equiv 1\pmod 4$. From $y\equiv 1\pmod 4$, $t\equiv 1\pmod 4$, and $E_{5}$, $2^{x} \equiv ( -1)^{z}\pmod 5$, which contradicts $x\equiv 1\pmod 2$. If $z=2$ and $t=1$, then $12^{x} +13^{y} =2209$, which is impossible. Assume that $z=2$ and $t\geq 2$. From $E_{9}$, we have $4^{y} +2\equiv 0\pmod 9$, which implise $y\equiv 2\pmod 3$. From $y\equiv 2\pmod 3$, $2013\equiv 0\pmod{61}$ and $E_{61}$, we have $12^{x} \equiv 27\pmod{61}$, impossible. So we must have $x=1$ and suppose henceforth that $x=1$. Assume that $z\geq 4$. From $E_{16}$ and $t\equiv 1\pmod 4$, we have $13^{y} \equiv 1\pmod{16}$, which implise $y\equiv 0\pmod 4$. From $t\equiv 1\pmod 4$, $y\equiv 0\pmod 4$, and $E_{5}$, we have $-14^{z} \equiv 0\pmod 5$, impossible. So we must have $z=2$ and suppose henceforth that $z=2$. Assume that $t\geq 2$. From $E_{9}$, we have $4^{y} \equiv 4\pmod 9$, which implise $y\equiv 1\pmod 3$. Since $y\equiv 1\pmod 3$, from $E_{61}$, we have $12\equiv 0\pmod{61}$, impossible. So we must have $t=1$ and suppose henceforth $t=1$. Finally, we have $13^{y} =2197$, which implise $y=3$. Hence the only solution for the given equation is $\boxed{( x,y,z,t) =( 1,3,2,1)}$.