Three, (50 points) Find all positive integer triples $(x, y, z)$ that satisfy $5^{x}+12^{y}=13^{z}$.

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Three, taking the given equation modulo 3 yields $(-1)^{x}=1(\bmod 3)$.
Thus, $2 \mid x$.
Let $x=2 x_{1}\left(x_{1} \in \mathbf{Z}_{+}\right)$.
(1) When $y=1$, the original equation becomes
$13^{z}-5^{2 x_{1}}=12$.
Taking both sides modulo 13 yields $(-1)^{x_{1}}=1(\bmod 13)$.
Thus, $2 \mid x_{1}$.
Let $x_{1}=2 x_{2}\left(x_{2} \in Z_{+}\right)$. Then the original equation transforms into
$13^{x}-5^{4 x_{2}}=12$.
Taking both sides modulo 5 yields $3^{x}=2(\bmod 5)$.
Thus, $z \equiv 3(\bmod 4)$.
Taking both sides modulo 16 yields
$$
5 \equiv(-3)^{x} \equiv 13^{x}=625^{x_{2}}+12 \equiv-3(\bmod 16),
$$

which is a contradiction.
(2) When $y \geqslant 2$, taking both sides of the original equation modulo 8 yields
$1 \equiv 5^{x}+12^{y}=13^{x} \equiv 5^{x}(\bmod 8)$.
Thus, $2 \mid z$.
Let $z=2 z_{1}\left(z_{1} \in Z_{+}\right)$. Then the original equation transforms into
$13^{2 x_{1}}-5^{2 x_{1}}=12^{y}$
$\Rightarrow\left(13^{x_{1}}-5^{x_{1}}\right)\left(13^{x_{1}}+5^{x_{1}}\right)=12^{y}$.
Notice that, $\left(13^{x_{1}}+5^{x_{1}}, 13^{x_{1}}-5^{x_{1}}\right)=2$.
Hence, $\left\{\begin{array}{l}13^{x_{1}}+5^{x_{1}}=2^{2 y-1}, \\ 13^{x_{1}}-5^{x_{1}}=2 \times 3^{y}\end{array}\right.$

or $\left\{\begin{array}{l}13^{x_{1}}+5^{x_{1}}=2 \times 3^{y}, \\ 13^{x_{1}}-5^{x_{1}}=2^{2 y-1} .\end{array}\right.$
From equation set (2), we get $5^{x_{1}}=2^{2(y-1)}-3^{y}$.
Taking both sides modulo 4 yields $(-1)^{y} \equiv-1(\bmod 4)$.
Thus, $y$ is odd.
Taking both sides modulo 5 yields
$$
3^{y} \equiv(-1)^{y-1}=1(\bmod 5),
$$

which is a contradiction.
Therefore, equation set (2) has no solution.
From equation set (3), we get $5^{x_{1}}=3^{y}-2^{2(y-1)}$.
Taking both sides modulo 4 yields $(-1)^{y} \equiv 1(\bmod 4)$.
Thus, $y$ is even.
Let $y=2 y_{1}\left(y_{1} \in \mathbf{Z}_{+}\right)$. Then the original equation transforms into
$3^{2 y_{1}}-2^{2\left(2 y_{1}-1\right)}=5^{x_{1}}$
$\Rightarrow\left(3^{y_{1}}-2^{2 y_{1}-1}\right)\left(3^{y_{1}}+2^{2 y_{1}-1}\right)=5^{x_{1}}$.
Notice that, $\left(3^{y_{1}}+2^{2 y_{1}-1}, 3^{y_{1}}-2^{2 y_{1}-1}\right)=1$.
Then $\left\{\begin{array}{l}3^{y_{1}}+2^{2 y_{1}-1}=5^{x_{1}}, \\ 3^{y_{1}}-2^{2 y_{1}-1}=1 .\end{array}\right.$
If $y_{1} \geqslant 2$, from $3^{y_{1}}-2^{2 y_{1}-1}=1$, taking both sides modulo 4 yields $(-1)^{y_{1}} \equiv 1(\bmod 4)$.
Thus, $2 \mid y_{1}$.
Let $y_{1}=2 y_{2}\left(y_{2} \in Z_{+}\right)$. Then the original equation transforms into $\left(3^{y_{2}}-1\right)\left(3^{y_{2}}+1\right)=2^{4 y_{2}-1}$.
Notice that, $\left(3^{y_{2}}-1,3^{y_{2}}+1\right)=2$.
Then $\left\{\begin{array}{l}3^{y_{2}}+1=2^{4 y_{2}-2} \\ 3^{y_{2}}-1=2\end{array} \Rightarrow y_{2}=1\right.$.
But at this point $y_{1}=2,5^{x_{1}}=3^{2}+2^{3}=17$, which is a contradiction.
Hence, $y_{1}=1, x_{1}=1, z_{1}=1$, i.e.,
$(x, y, z)=(2,2,2)$.