Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.
1. We start with the number \(11111\) in base \(n\). This can be expressed as:
   \[
   n^4 + n^3 + n^2 + n + 1
   \]
   We need to determine for which integers \(n \geq 2\) this expression is a perfect square.

2. Let \(m\) be an integer such that:
   \[
   m^2 = n^4 + n^3 + n^2 + n + 1
   \]

3. To find \(m\), we consider the inequality:
   \[
   (2n^2 + n)^2 < 4n^4 + 4n^3 + 4n^2 + 4n + 4 < (2n^2 + n + 2)^2
   \]

4. Expanding the squares, we get:
   \[
   (2n^2 + n)^2 = 4n^4 + 4n^3 + n^2
   \]
   \[
   (2n^2 + n + 2)^2 = 4n^4 + 4n^3 + 4n^2 + 4n + 4
   \]

5. We see that:
   \[
   4n^4 + 4n^3 + 4n^2 + 4n + 4 = (2n^2 + n + 1)^2
   \]

6. Equating the expressions, we get:
   \[
   4n^4 + 4n^3 + 4n^2 + 4n + 4 = (2n^2 + n + 1)^2
   \]

7. Expanding the right-hand side:
   \[
   (2n^2 + n + 1)^2 = 4n^4 + 4n^3 + n^2 + 4n^2 + 2n + 1
   \]
   \[
   = 4n^4 + 4n^3 + 5n^2 + 2n + 1
   \]

8. Setting the expanded form equal to the left-hand side:
   \[
   4n^4 + 4n^3 + 4n^2 + 4n + 4 = 4n^4 + 4n^3 + 5n^2 + 2n + 1
   \]

9. Simplifying, we get:
   \[
   4n^2 + 2n + 3 = 0
   \]

10. Solving the quadratic equation:
    \[
    n^2 - 2n - 3 = 0
    \]
    Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -3\):
    \[
    n = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}
    \]
    \[
    n = 3 \quad \text{or} \quad n = -1
    \]

11. Since \(n \geq 2\), we discard \(n = -1\) and keep \(n = 3\).

Conclusion:
\[
\boxed{n = 3}
\]