Find all pairs of positive integers $m, n \ge 3$ for which there exist infinitely many positive integers $a$ such that \[\frac{a^{m}+a-1}{a^{n}+a^{2}-1}\] is itself an integer.
1. We start by assuming that there are infinitely many positive integers \( a \) such that the expression \(\frac{a^m + a - 1}{a^n + a^2 - 1}\) is an integer. This implies that \( a^n + a^2 - 1 \) divides \( a^m + a - 1 \).

2. Since \( m \) and \( n \) are positive integers with \( m, n \ge 3 \), we can assume without loss of generality that \( m > n \). Let \( m = n + k \) for some \( k > 0 \).

3. We rewrite the expression \( a^m + a - 1 \) in terms of \( n \) and \( k \):
   \[
   a^m + a - 1 = a^{n+k} + a - 1
   \]

4. We need to show that \( a^n + a^2 - 1 \) divides \( a^{n+k} + a - 1 \). We can express this as:
   \[
   a^{n+k} + a - 1 = (a^n + a^2 - 1)Q(a) + R(a)
   \]
   where \( Q(a) \) is the quotient and \( R(a) \) is the remainder when \( a^{n+k} + a - 1 \) is divided by \( a^n + a^2 - 1 \).

5. Since \( a^n + a^2 - 1 \) divides \( a^{n+k} + a - 1 \) for infinitely many \( a \), the remainder \( R(a) \) must be zero for these values of \( a \). Therefore, \( R(a) \) must be identically zero, implying that \( a^n + a^2 - 1 \) divides \( a^{n+k} + a - 1 \) exactly.

6. We now consider the polynomial division:
   \[
   a^{n+k} + a - 1 = (a^n + a^2 - 1)Q(a)
   \]
   To find \( Q(a) \), we need to match the degrees of the polynomials. The degree of \( a^{n+k} \) is \( n+k \), and the degree of \( a^n + a^2 - 1 \) is \( n \). Therefore, \( Q(a) \) must be a polynomial of degree \( k \).

7. Let \( Q(a) = a^k + c_{k-1}a^{k-1} + \cdots + c_0 \). Then:
   \[
   a^{n+k} + a - 1 = (a^n + a^2 - 1)(a^k + c_{k-1}a^{k-1} + \cdots + c_0)
   \]

8. Expanding the right-hand side and equating coefficients, we get:
   \[
   a^{n+k} + a - 1 = a^{n+k} + c_{k-1}a^{n+k-1} + \cdots + c_0a^n + a^{k+2} + c_{k-1}a^{k+1} + \cdots + c_0a^2 - a^k - c_{k-1}a^{k-1} - \cdots - c_0
   \]

9. For the polynomial identity to hold, the coefficients of corresponding powers of \( a \) must match. This implies that the terms involving \( a \) and the constant term must cancel out. This leads to the system of equations:
   \[
   \begin{cases}
   c_0 = 1 \\
   c_{k-1} = 0 \\
   \vdots \\
   c_1 = 0
   \end{cases}
   \]

10. Solving these equations, we find that \( k = 2 \), \( n = 3 \), and \( m = n + k = 5 \).

11. Finally, we verify that the identity holds:
   \[
   a^5 + a - 1 = (a^3 + a^2 - 1)(a^2 - a + 1)
   \]
   Expanding the right-hand side:
   \[
   (a^3 + a^2 - 1)(a^2 - a + 1) = a^5 - a^4 + a^3 + a^4 - a^3 + a^2 - a^2 + a - 1 = a^5 + a - 1
   \]
   This confirms that the identity is correct.

The final answer is \( \boxed{ (m, n) = (5, 3) } \).