If \[2011^{2011^{2012}} = x^x\] for some positive integer $x$, how many positive integer factors does $x$ have?

[i]Author: Alex Zhu[/i]
1. We start with the given equation:
   \[
   2011^{2011^{2012}} = x^x
   \]
   We need to find a positive integer \( x \) that satisfies this equation.

2. Since \( 2011 \) is a prime number, \( x \) must be a power of \( 2011 \). Let \( x = 2011^a \) for some integer \( a \).

3. Substituting \( x = 2011^a \) into the equation, we get:
   \[
   2011^{2011^{2012}} = (2011^a)^{2011^a}
   \]

4. Simplifying the right-hand side, we have:
   \[
   (2011^a)^{2011^a} = 2011^{a \cdot 2011^a}
   \]

5. Equating the exponents of \( 2011 \) on both sides, we get:
   \[
   2011^{2012} = a \cdot 2011^a
   \]

6. To solve for \( a \), we observe that the exponents must match. Therefore, we set:
   \[
   a \cdot 2011^a = 2011^{2012}
   \]

7. Since the bases are the same, we can equate the exponents:
   \[
   a \cdot 2011^a = 2011^{2012}
   \]

8. Clearly, \( a = 2011 \) satisfies this equation because:
   \[
   2011 \cdot 2011^{2011} = 2011^{2012}
   \]

9. Now, we need to find the number of positive integer factors of \( x \). Since \( x = 2011^{2011} \) and \( 2011 \) is a prime number, the number of factors of \( 2011^{2011} \) is given by:
   \[
   2011^{2011} \text{ has } 2011 + 1 = 2012 \text{ factors}
   \]

The final answer is \(\boxed{2012}\).