Find all triples of positive integers $(a, m, n)$ satisfying $a \geqslant 2, m \geqslant 2$ such that $a^{n}+203$ is divisible by $a^{m}+1$.

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Find all triples of positive integers $(a, m, n)$ satisfying $a \geqslant 2, m \geqslant 2$ such that $a^{n}+203$ is divisible by $a^{m}+1$.

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For $n, m$, we discuss in three cases.
(i) When $nm$, from $a^{m}+1 \mid 203\left(a^{m}+1\right)$, we have $a^{m}+11 a^{n}+203-\left(203 a^{m}+203\right)$ $=a^{m}\left(a^{n-m}-203\right)$.
Also, $\left(a^{m}+1, a^{m}\right)=1$, so $a^{m}+1 \mid a^{n-m}-203$.
(1) If $a^{n-m}203$, let $n-m=s \geqslant 1$, then
$$
a^{m}+11 a^{s}-203 \text {. }
$$

Also, $a^{s}-203 \geqslant a^{m}+1$, so $s>m$.
From $a^{m}+11 a^{s}+203 a^{m}=\left(a^{s-m}+203\right) a^{m}$ $=\left(a^{n-2 m}+203\right) a^{m}$,
$\left(a^{m}+1, a^{m}\right)=1$,
so $a^{m}+11 a^{n-2 m}+203$.
Also, $s>m \Leftrightarrow n-m>m \Leftrightarrow n>2 m \Leftrightarrow n-2 m>0$.
The solutions at this point can only be derived from the previous solutions, i.e.,
$$
(a, m, n) \rightarrow(a, m, n+2 m) \rightarrow \cdots \rightarrow(a, m, n+2 k m) \text {, }
$$

and each derived solution satisfies $a^{m}+11 a^{n}+203$.
In summary, all solutions $(a, m, n)$ are
$$
\begin{array}{l}
(2,2,4 k+1),(2,3,6 k+2),(2,4,8 k+8), \\
(2,6,12 k+9),(3,2,4 k+3),(4,2,4 k+4), \\
(5,2,4 k+1),(8,2,4 k+3),(10,2,4 k+2), \\
(203, m,(2 k+1) m+1),
\end{array}
$$

where $k$ is any non-negative integer, and $m \geqslant 2$ is an integer.