Find all triples $(x, y, z)$ of positive integers such that 
\[x^2 + 4^y = 5^z. \]

[i]Proposed by Li4 and ltf0501[/i]
We are given the equation:
\[ x^2 + 4^y = 5^z \]
and we need to find all triples \((x, y, z)\) of positive integers that satisfy this equation.

1. **Case 1: \( y > 1 \)**

   First, consider the equation modulo 8:
   \[ 5^z \equiv x^2 \pmod{8} \]
   Since \(5 \equiv 5 \pmod{8}\), we have:
   \[ 5^z \equiv 5 \pmod{8} \]
   The quadratic residues modulo 8 are \(0, 1, 4\). Therefore, \(x^2 \equiv 1 \pmod{8}\), which implies:
   \[ 5^z \equiv 1 \pmod{8} \]
   This is only possible if \(z\) is even. Let \(z = 2z'\). Substituting \(z = 2z'\) into the original equation, we get:
   \[ x^2 + 4^y = 5^{2z'} \]
   This can be factored as:
   \[ (5^{z'} - x)(5^{z'} + x) = 4^y \]
   Since \(4^y = 2^{2y}\), we have:
   \[ \gcd(5^{z'} - x, 5^{z'} + x) \mid \gcd(2 \cdot 5^{z'}, 2^{2y}) = 2 \]
   Therefore, the factors \(5^{z'} - x\) and \(5^{z'} + x\) must be \(2\) and \(2^{2y-1}\) respectively. This gives us:
   \[ 5^{z'} - x = 2 \]
   \[ 5^{z'} + x = 2^{2y-1} \]
   Adding these two equations, we get:
   \[ 2 \cdot 5^{z'} = 2 + 2^{2y-1} \]
   \[ 5^{z'} = 1 + 2^{2y-2} \]
   By Mihăilescu's theorem (also known as Catalan's conjecture), the only solution in natural numbers to the equation \(x^a - y^b = 1\) for \(a, b > 1\) is \(3^2 - 2^3 = 1\). Therefore, the only solution is:
   \[ 5^{z'} = 5 \]
   \[ z' = 1 \]
   Substituting \(z' = 1\) back, we get:
   \[ x = 3 \]
   \[ 2^{2y-2} = 4 \]
   \[ 2y - 2 = 2 \]
   \[ y = 2 \]
   Thus, we have the solution \((x, y, z) = (3, 2, 2)\).

2. **Case 2: \( y = 1 \)**

   If \(y = 1\), the equation becomes:
   \[ x^2 + 4 = 5^z \]
   Rearranging, we get:
   \[ x^2 = 5^z - 4 \]
   We need to check if \(5^z - 4\) is a perfect square. Let's test small values of \(z\):
   - For \(z = 1\):
     \[ 5^1 - 4 = 1 \]
     \[ x = 1 \]
     Thus, \((x, y, z) = (1, 1, 1)\) is a solution.
   - For \(z = 2\):
     \[ 5^2 - 4 = 21 \]
     21 is not a perfect square.
   - For \(z = 3\):
     \[ 5^3 - 4 = 121 \]
     \[ x = 11 \]
     Thus, \((x, y, z) = (11, 1, 3)\) is a solution.

   For \(z > 3\), \(5^z - 4\) grows rapidly and is unlikely to be a perfect square. We can use properties of Pell's equation to show that there are no further solutions.

The final answer is \( \boxed{ (x, y, z) = (3, 2, 2) } \), \((1, 1, 1)\), and \((11, 1, 3)\).