KMO Problem # 1

WLOG, $x \ge y$

Since $1 + 4^x + 4^y$ is odd, $\implies z = 2k + 1 ; k \in \ {0, 1, 2, ...}$

$1 + 4^x + 4^y = z^2 = (2k + 1)^2 \\ 1 + 2^{2x} + 2^{2y} = 4k(k+1) + 1 \\ 2^{2x} + 2^{2y} = 4k(k + 1) \\ 2^{2x - 2} + 2^{2y - 2} = k(k+1)$

let $k = 2^v \implies z = 2^{v + 1} + 1$ $2^{2x - 2} + 2^{2y - 2} = \left( 2^v \right) \left( 2^v + 1\right) = 2^{2v} + 2^v \\ 2x - 2 = 2v , 4y - 4 = 2v \\ x = v + 1, y = \cfrac{2v + 4}{4} = \ \cfrac{v+2}{2}$

v must be even $\implies v = 2p \implies z = 2^{2p + 1} + 1$ $x = 2p + 1, y = p + 1, z = 2^{2p + 1} + 1 \\ 2^{2p + 1} + 1 \le 10,000 \\ 2^{2p + 1} \le 9,999 \\ 2p + 1 \le log_{2} 9999 \approx 13.28 \\ \text{max} \ (2p + 1) = 13 \\ \text{max} \ (z \le 10,000) = 2^{13} + 1 = \ \boxed{8,193}$