Find all functions $f: \mathbb N \cup \{0\} \to \mathbb N\cup \{0\}$ such that $f(1)>0$ and 
\[f(m^2+3n^2)=(f(m))^2 + 3(f(n))^2 \quad \forall m,n \in \mathbb N\cup \{0\}.\]
1. Given the functional equation:
   \[
   f(m^2 + 3n^2) = (f(m))^2 + 3(f(n))^2 \quad \forall m, n \in \mathbb{N} \cup \{0\}
   \]
   and the condition \( f(1) > 0 \).

2. First, consider the case when \( m = n = 0 \):
   \[
   f(0^2 + 3 \cdot 0^2) = (f(0))^2 + 3(f(0))^2 \implies f(0) = 4(f(0))^2
   \]
   Let \( f(0) = c \). Then:
   \[
   c = 4c^2 \implies 4c^2 - c = 0 \implies c(4c - 1) = 0
   \]
   This gives us two solutions: \( c = 0 \) or \( c = \frac{1}{4} \).

3. Since \( f: \mathbb{N} \cup \{0\} \to \mathbb{N} \cup \{0\} \), \( f(0) \) must be a non-negative integer. Therefore, \( f(0) = \frac{1}{4} \) is not possible. Hence, \( f(0) = 0 \).

4. Now, substitute \( n = 0 \) in the original equation:
   \[
   f(m^2 + 3 \cdot 0^2) = (f(m))^2 + 3(f(0))^2 \implies f(m^2) = (f(m))^2
   \]
   This implies that \( f \) is multiplicative for squares.

5. Consider \( m = 1 \):
   \[
   f(1^2) = (f(1))^2 \implies f(1) = (f(1))^2
   \]
   Since \( f(1) > 0 \), the only solution is \( f(1) = 1 \).

6. Next, consider \( m = 2 \):
   \[
   f(2^2) = (f(2))^2 \implies f(4) = (f(2))^2
   \]
   We need to determine \( f(2) \). Consider \( m = 2 \) and \( n = 0 \):
   \[
   f(2^2 + 3 \cdot 0^2) = (f(2))^2 + 3(f(0))^2 \implies f(4) = (f(2))^2
   \]
   Since \( f(4) = 4 \), we have:
   \[
   4 = (f(2))^2 \implies f(2) = 2
   \]

7. Now, consider \( m = 3 \):
   \[
   f(3^2) = (f(3))^2 \implies f(9) = (f(3))^2
   \]
   We need to determine \( f(3) \). Consider \( m = 3 \) and \( n = 0 \):
   \[
   f(3^2 + 3 \cdot 0^2) = (f(3))^2 + 3(f(0))^2 \implies f(9) = (f(3))^2
   \]
   Since \( f(9) = 9 \), we have:
   \[
   9 = (f(3))^2 \implies f(3) = 3
   \]

8. Assume \( f(k) = k \) for all \( k \in \mathbb{N} \). We need to show that this assumption holds for all \( k \). Consider \( m = k \) and \( n = 0 \):
   \[
   f(k^2 + 3 \cdot 0^2) = (f(k))^2 + 3(f(0))^2 \implies f(k^2) = (f(k))^2
   \]
   Since \( f(k) = k \), we have:
   \[
   f(k^2) = k^2
   \]
   This implies that \( f(k) = k \) for all \( k \in \mathbb{N} \).

9. Therefore, the only function that satisfies the given functional equation is:
   \[
   f(n) = n \quad \forall n \in \mathbb{N} \cup \{0\}
   \]

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