Mongolian Mathematical Olympiad

From the contrary, suppose there exist integers $(a, b)$ with $\gcd(a, b) = 1$ and $|a| > |b| > 1$ that satisfy the equation
$$(a^n - b^n)^2 = a^{n+m} - b^{n+m}$$
for $n \ge m \ge 1$.
We can rewrite the given equation as follows:
$$(a^n - b^n)^2 = a^n(a^m - b^m) + b^m(a^n - b^n).$$
This implies that $a^n - b^n$ divides $a^m - b^m$. Let $S = \frac{a^m - b^m}{a^n - b^n}$. Then, dividing the equation by $a^n - b^n$, we have:
$$a^n - b^n = a^n S + b^m,$$
that simplifies to:
$$a^n(1 - S) = b^m(b^{n-m} + 1).$$
Since $|b| > 1$ and $b^{n-m} + 1 \ne 0$, we conclude that $a^n$ divides $b^{n-m} + 1$. Thus,
$$a^n \mid b^{n-m} + 1.$$
Now, consider the inequality:
$$|b|^{n} + 1 \le (|b| + 1)^{n} \le |a|^{n} \le |b|^{n-m} + 1.$$
This implies: $|b|^{n} + 1 < |b|^{n} + 1$, which is a contradiction.
Hence, there are no pairs $(a, b)$ that satisfy the conditions of the problem.