Taiwan IMO Selection Camp, 2021

All solutions are $(1, 1, 1)$, $(11, 1, 3)$, $(3, 2, 2)$.

We first deal with the case $y \ge 2$. We have
$$x^2 \equiv 5^z \pmod{8},$$
and $x^2 \equiv 0, 1, 4$, $5^z \equiv 1, 5$, so $z$ is even. Let $z = 2z'$, then
$$x^2 + (2^y)^2 = (5^{z'})^2.$$
Since $5^{z'}$ and $2^y$ are coprime and $2 \mid 2^y$, by the Pythagorean triple formula we have
$$x = m^2 - n^2, \quad 2^y = 2mn, \quad 5^{z'} = m^2 + n^2,$$
where $m, n$ are coprime and $m > n$. From $2^y = 2mn$ we get $m = 2^{y-1}, n = 1$, so $5^{z'} = 2^{2(y-1)}+1$.
If $y \ge 3$, then
$$5^{z'} \equiv 1 \pmod{8} \implies 2 \mid z',$$
let $z' = 2z''$, then
$$(5^{z''} + 1)(5^{z''} - 1) = 2^{2(y-1)} \implies 5^{z''} + 1 = 2^a,\ 5^{z''} - 1 = 2^b.$$
Note that $4$ does not divide one of $5^{z''} + 1$ and $5^{z''} - 1$, so $a = 1$ or $b = 1$, and it is easy to see there is no solution in this case.
If $y = 2$, we obtain the solution $(x, y, z) = (3, 2, 2)$.

Now suppose $y = 1$. We have
$$x^2 + 4 \equiv 5^z \pmod{8},$$
and $x^2 + 4 \equiv 0, 4, 5$, $5^z \equiv 1, 5$, so $z$ is odd. Let $z = 2z' + 1$, and consider the Pell equation
$$s^2 - 5t^2 = -4,$$
we wish to find solutions $(s, t) = (x, 5^{z'})$. If $(s, t)$ is a solution in positive integers, we know that $(\frac{3s-5t}{2}, \frac{3t-s}{2})$ is also a solution in integers (note that $s, t$ have the same parity), so we can always descend a solution $(s, t)$ until
$$3s \le 5t \text{ or } 3t \le s.$$
Note that $3t > s$ always holds. For $3s \le 5t$, we have
$$-20t^2 = 25t^2 - 45t^2 \ge 9(s^2 - 5t^2) = -36 \implies t = 1 \implies s = 1.$$
Thus we obtain all positive integer solutions $(s_n, t_n)$ satisfying
$$(s_0, t_0) = (1, 1), \quad s_{n+1} = \frac{3s_n + 5t_n}{2}, \quad t_{n+1} = \frac{3t_n + s_n}{2}.$$
Observing the first few terms $(s_1, t_1) = (4, 2)$, $(s_2, t_2) = (11, 5)$, ..., and considering the Fibonacci sequence
$$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, \dots,$$
from the above recurrence we can guess and verify that
$$(s_n, t_n) = (F_{2n+2} + F_{2n}, F_{2n+1}).$$
If $t_n = 5^{z'}$, we get $z' = 0$ or $5 \mid F_{2n+1}$. The former gives the solution $(x, y, z) = (1, 1, 1)$, and the latter, by the pattern of the Fibonacci sequence modulo 5, gives $5 \mid n' := 2n + 1$. If $p \ne 5$ is a prime factor of $n'$, then $p > 2$ and
$$F_p \mid F_{n'} = 5^{z'},$$
but $F_p \ne 1$ and $5 \nmid F_p$, a contradiction. Therefore $n'$ is a power of 5. If $z' > 1$, then $n' > 5$, so $25 \mid n'$, hence $F_{25} \mid F_{n'} = 5^{z'}$, but $F_{25} = 75025$ is not a power of 5, a contradiction. Therefore $z' = 1$, and we obtain the solution $(x, y, z) = (11, 1, 3)$.