Find all positive integers $n$ such that $n^4 - n^3 + 3n^2 + 5$ is a perfect square.
To find all positive integers \( n \) such that \( n^4 - n^3 + 3n^2 + 5 \) is a perfect square, we will consider two cases: when \( n \) is even and when \( n \) is odd.

### Case 1: \( n \) is even
1. Let \( n = 2k \).
2. We need to check if \( n^4 - n^3 + 3n^2 + 5 \) can be written as \( m^2 \) for some integer \( m \).
3. Consider the expression \( (n^2 - k + 1)^2 \):
   \[
   (n^2 - k + 1)^2 = n^4 - 2kn^2 + k^2 + 2n^2 - 2k + 1
   \]
4. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   n^4 - n^3 + 3n^2 + 5 < (n^2 - k + 1)^2
   \]
   Simplifying, we get:
   \[
   0 < \frac{7n^2}{4} + n + 4 \quad \text{(TRUE)}
   \]
5. Now consider \( (n^2 - k + 2)^2 \):
   \[
   (n^2 - k + 2)^2 = n^4 - 2kn^2 + k^2 + 4n^2 - 4k + 4
   \]
6. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   (n^2 - k + 2)^2 > n^4 - n^3 + 3n^2 + 5
   \]
   Simplifying, we get:
   \[
   \frac{5n^2}{4} - 2n - 4 > 0 \quad \Rightarrow n \ge 3
   \]
7. Testing \( n = 2 \):
   \[
   2^4 - 2^3 + 3 \cdot 2^2 + 5 = 16 - 8 + 12 + 5 = 25 \quad \text{(which is a perfect square)}
   \]
   Therefore, \( n = 2 \) is a solution.

### Case 2: \( n \) is odd
1. Let \( n = 2k + 1 \).
2. We need to check if \( n^4 - n^3 + 3n^2 + 5 \) can be written as \( m^2 \) for some integer \( m \).
3. Consider the expression \( (n^2 - k)^2 \):
   \[
   (n^2 - k)^2 = n^4 - 2kn^2 + k^2
   \]
4. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   n^4 - n^3 + 3n^2 + 5 > (n^2 - k)^2
   \]
   Simplifying, we get:
   \[
   0 < \frac{3n^2}{2} + n + \frac{9}{2} \quad \text{(TRUE)}
   \]
5. Now consider \( (n^2 - k + 1)^2 \):
   \[
   (n^2 - k + 1)^2 = n^4 - 2kn^2 + k^2 + 2n^2 - 2k + 1
   \]
6. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   (n^2 - k + 1)^2 > n^4 - n^3 + 3n^2 + 5
   \]
   Simplifying, we get:
   \[
   5n^2 - 6n - 11 > 0 \quad \Rightarrow n \ge 3
   \]
7. Testing \( n = 1 \):
   \[
   1^4 - 1^3 + 3 \cdot 1^2 + 5 = 1 - 1 + 3 + 5 = 8 \quad \text{(which is not a perfect square)}
   \]
   Therefore, there are no solutions for odd \( n \).

### Conclusion:
The only solution is \( n = 2 \).

The final answer is \( \boxed{ n = 2 } \)