Suppose $ x + y + z = a^2$, $ a\in \mathbb{N}$. We have $ x^2 + y^2 + z^2 < (x + y + z)^2\le 3(x^2 + y^2 + z^2)$, that is $ 1993 < a^4 < 2\cdot 1993$, so $ 6 < a < 9$. Because $ a^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$ and $ x^2 + y^2 + z^2 = 1993$ is odd, we obtain that $ a$ is odd too, so $ a = 7$. $ x + y + z = 49$, $ y + z = 49 - x$ and $ (y + z)^2\geq y^2 + z^2$, that is $ (49 - x)^2\geq 1993 - x^2$ or $ x^2 - 49x + 204\geq0$. We obtain that $ x\geq \dfrac{49 + \sqrt {1585}}{2}$ or $ x\le \dfrac{49 - \sqrt {1585}}{2}$. In first case $ x > 45$, so $ x^2 > 1993 = x^2 + y^2 + z^2$. In the secod case $ x\le 4$, similiarly $ y \le 4, z\le 4$, contradiction....