Korean Mathematical Olympiad Final Round

Without loss of generality, we may assume that $x \le y$. Suppose that $2x < y + 1$. Then
$$(2^y)^2 < 1 + 4^x + 4^y < (1 + 2^y)^2,$$
which implies that $1 + 4^x + 4^y$ is not a square of an integer. If $2x = y + 1$, then $1 + 4^x + 4^y = 1 + 2^{y+1} + 4^y = (1 + 2^y)^2$. Hence
$$(x, y, z) = (x, 2x-1, 1 + 2^{2x-1})$$
is a solution of the equation for any positive integer $x$. Suppose that $2x > y + 1$. Note that
$$4^x + 4^y = 4^x(1 + 4^{y-x}) = (z-1)(z+1).$$
Since $\gcd(z-1, z+1) = 2$, one of $z-1$ or $z+1$ is divisible by $2^{2x-1}$. This gives a contradiction because
$$2(1 + 4^{y-x}) \le 2(1 + 4^{x-2}) < 2^{2x-1} - 2$$
for any $x > 1$. Therefore the solutions are:
$$(x, y, z) = (x, 2x-1, 1 + 2^{2x-1}) \text{ or } (2x-1, x, 1 + 2^{2x-1})$$
for any positive integer $x$. $\square$