We will discuss the following three cases for $n$ and $m$.

(i) In the case when $n < m$, from $a^n + 203 \ge a^m + 1$, we have
$$202 \ge a^m - a^n \ge a^n(a-1) \ge a(a-1).$$
Therefore, $2 \le a \le 14$.

When $a = 2$, we can take $n$ to be $1, 2, \dots, 7$.
When $a = 3$, we can take $n$ to be $1, 2, 3$ and $4$.
When $a = 4$, we can take $n$ to be $1, 2$ and $3$.
When $5 \le a \le 6$, we can take $n$ to be $1$ and $2$.
When $7 \le a \le 14$, $n = 1$.
From $a^m + 1 \mid a^n + 203$, we can deduce that the solutions are $(2, 2, 1), (2, 3, 2)$ and $(5, 2, 1)$.

(ii) In the case when $n = m$, $a^m + 1 \mid 202$. Since 202 has only four divisors 1, 2, 101 and 202, but $a \ge 2$, $m \ge 2$ and $a^m + 1 \ge 5$, so $a^m = 100$ or 201. Since $m \ge 2$, therefore the solution is $(10, 2, 2)$.

(iii) In the case when $n > m$, from $a^m + 1 \mid 203(a^m + 1)$, we have
$$a^m + 1 \mid a^n + 203 - (203a^m + 203),$$
that is,
$$a^m + 1 \mid a^m(a^{n-m} - 203).$$
Since $(a^m + 1, a^m) = 1$, so
$$a^m + 1 \mid a^{n-m} - 203.$$

① If $a^{n-m} < 203$, then set $n-m = s \ge 1$, we have $a^m + 1 \mid 203-a^s$. Hence
$$203 - a^s \ge a^m + 1, \\ 202 \ge a^s + a^m \ge a^m + a \\ = a(a^{m-1} + 1) \ge a(a+1),$$
thus
$$2 \le a \le 13.$$
Using the same argument as in Case (i), we show that the solutions for $(a, m, s)$ are:
$(2, 2, 3), (2, 6, 3), (2, 4, 4), (2, 3, 5), (2, 2, 7), (3, 2, 1), (4, 2, 2), (5, 2, 3)$ and $(8, 2, 1)$.
Hence, $(a, m, n)$ are
$(2, 2, 5), (2, 6, 9), (2, 4, 8), (2, 3, 8), (2, 2, 9), (3, 2, 3), (4, 2, 4), (5, 2, 5)$ and $(8, 2, 3)$.

② If $a^{n-m} = 203$, then $a = 203$ and $n-m = 1$, and the solution is
$(203, m, m+1)$, $m \ge 2$.

③ If $a^{n-m} > 203$, set $n-m = s \ge 1$, then $a^m + 1 \mid a^s - 203$.
Since $a^s - 203 \ge a^m + 1$, so $s > m$. By
$$a^m + 1 \mid a^s + 203a^m = (a^{s-m} + 203)a^m \\ = (a^{n-2m} + 203)a^m,$$
and
$(a^m + 1, a^m) = 1,$
we have
$a^m + 1 \mid a^{n-2m} + 203.$
Now $s > m \Leftrightarrow n - m > m \Leftrightarrow n > 2m \Leftrightarrow n - 2m > 0$. In this case, the solutions can only be derived from the preceding solutions, that is, from $(a, m, n) \rightarrow (a, m, n+2m) \rightarrow \dots \rightarrow (a, m, n+2km)$, and each solution derived also satisfies $a^m + 1 \mid a^n + 203$.

Summarizing what described above, we obtain all solutions $(a, m, n)$ to be:
$(2, 2, 4k+1)$, $(2, 3, 6k+2)$, $(2, 4, 8k+8)$, $(2, 6, 12k+9)$,
$(3, 2, 4k+3)$, $(4, 2, 4k+4)$, $(5, 2, 4k+1)$, $(8, 2, 4k+3)$,
$(10, 2, 4k+2)$ and $(203, m, (2k+1)m+1)$,
where $k$ is any nonnegative integer, and $m \ge 2$ is an integer.