Solution It is not difficult to verify that $ ( x,y) =( 3,1)$ is a solution to the equation. We want to show that there is no other solution to the equation. Assume that $ 2^{x} +19^{y} =z^{3}$ ... $ ( \heartsuit )$ for some $ ( x,y,z) \in \mathbb{Z}^{3}_{ >0}$ with $ x\neq 3$. Considering $ ( \heartsuit )$ modulo $ 9$, we have $ 2^{x} +1\equiv z^{3}\pmod 9$. It follows that $ x\equiv 3,4\pmod 6$. We consider the following two cases (1)(2) : (1) $ x\equiv 4\pmod 6$ Considering $ ( \heartsuit )$ modulo $ 19$, we have $ 2^{x} \equiv z^{3}\pmod{19}$. It is not difficult to verify that this is impossible. (2) $ x\equiv 3\pmod 6$ We can take $ n\in \mathbb{Z}$ such that $ x=3n$ and $ n >1$. Then $ 19^{y} =\left( z-2^{n}\right)\left( z^{2} +z2^{n} +2^{2n}\right)$. Since $ \text{gcd}\left( z-2^{n} ,z^{2} +z2^{n} +2^{2n}\right) =1$, we have $ z-2^{n} =1$ and $ z^{2} +z2^{n} +2^{2n} =19^{y}$. This implies that $ 3\cdotp 2^{n}\left( 2^{n} +1\right) =19^{y} -1$. Since $ n >1$, we have $ 4\mid 19^{n} -1$, which implies $ 2\mid y$. Since $ 9\mid 19 -1$ , we must have $ 3\mid 2^{n} +1$, which implise $ 2\nmid n$. Since $ 5\mid 19^{2} -1$ , we must have $ 5\mid 2^{n} +1$, which implise $ 2\mid n$. This is clearly a contradiction. $ \blacksquare $