Find all functions $f:\mathbb{Z}_{>0}\rightarrow\mathbb{Z}_{>0}$ that satisfy the following two conditions:
[list]$\bullet\ f(n)$ is a perfect square for all $n\in\mathbb{Z}_{>0}$
$\bullet\ f(m+n)=f(m)+f(n)+2mn$ for all $m,n\in\mathbb{Z}_{>0}$.[/list]
1. We are given that \( f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0} \) satisfies two conditions:
   - \( f(n) \) is a perfect square for all \( n \in \mathbb{Z}_{>0} \).
   - \( f(m+n) = f(m) + f(n) + 2mn \) for all \( m, n \in \mathbb{Z}_{>0} \).

2. Since \( f(n) \) is a perfect square, we can write \( f(n) = g(n)^2 \) where \( g(n) \) is a positive integer.

3. Substituting \( f(n) = g(n)^2 \) into the functional equation, we get:
   \[
   g(m+n)^2 = g(m)^2 + g(n)^2 + 2mn
   \]

4. Notice that the right-hand side can be rewritten using the identity \( (a+b)^2 = a^2 + b^2 + 2ab \):
   \[
   g(m+n)^2 = g(m)^2 + g(n)^2 + 2g(m)g(n) + 2mn - 2g(m)g(n)
   \]
   Simplifying, we get:
   \[
   g(m+n)^2 = (g(m) + g(n))^2 + 2mn - 2g(m)g(n)
   \]

5. For the equation to hold for all \( m, n \in \mathbb{Z}_{>0} \), we must have:
   \[
   2mn - 2g(m)g(n) = 0
   \]
   which simplifies to:
   \[
   mn = g(m)g(n)
   \]

6. Since \( g(n) \) is strictly increasing and positive, we can infer that \( g(n) = n \). This is because the only function that satisfies \( mn = g(m)g(n) \) for all \( m, n \in \mathbb{Z}_{>0} \) is \( g(n) = n \).

7. Substituting \( g(n) = n \) back into \( f(n) = g(n)^2 \), we get:
   \[
   f(n) = n^2
   \]

8. To verify, we substitute \( f(n) = n^2 \) into the original functional equation:
   \[
   f(m+n) = (m+n)^2 = m^2 + n^2 + 2mn = f(m) + f(n) + 2mn
   \]
   which holds true.

Therefore, the unique solution is \( f(n) = n^2 \).

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