$x^2 \equiv 5^y + 3^z \equiv (-1)^y \pmod{3}$ So, $y$ is even and let $y = 2y_0 \implies$ $x^2 - 5^y = (x - 5^{y_0})(x + 5^{y_0}) = 3^z$ $\implies \begin{cases} x + 5^{y_0} = 3^a \\ x - 5^{y_0} = 3^b \end{cases}$ where $z = a + b$. Then, $2\cdot 5^{y_0} = 3^a - 3^b$ how $a > b$ we get $b = 0$ (otherwise $3 \mid 2\cdot 5^{y_0} \to$ $Abs!$) and $2 \cdot 5^{y_0} = 3^a - 1$. If $a$ is even $4 \mid 3^a - 1 = 2\cdot 5^{y_0} \to Abs!$ thus $a$ is odd but $5 \mid 3^a - 1$ $\implies ord_5(3) = 4 \mid a \to$ $Abs!$ $\therefore \nexists$ $(x, y , z)$

mod 5 implies z even mod 4 implies z odd

First take mod 5 and get $3^z$ is a QR mod 5 so z has to be even (since $3$ is not a QR mod 5). Now if we take mod 4, the RHS becomes 1 + 1 = 2 mod 4, not a QR, absurd

I believe that using mod 5 and mod 4 solves the problem