Step 1: Concept
A polynomial of degree 2 or 3 is irreducible over a field if and only if it has no roots in that field.
Step 2: Meaning
The field $Z_3$ contains $\{0, 1, 2\}$. We test these values in $x^2 + 1$.
Step 3: Analysis
For $f(x) = x^2 + 1$:
$f(0) = 0^2 + 1 = 1 \ne 0$.
$f(1) = 1^2 + 1 = 2 \ne 0$.
$f(2) = 2^2 + 1 = 5 \equiv 2 \ne 0$.
Since none of the elements in $Z_3$ are roots, $x^2 + 1$ cannot be factored into linear terms.
Step 4: Conclusion
Therefore, $x^2 + 1$ is irreducible over $Z_3$.
Final Answer: (C)