Find all nonzero subsets $A$ of the set $\left\{2,3,4,5,\cdots\right\}$ such that $\forall n\in A$, we have that $n^{2}+4$ and $\left\lfloor{\sqrt{n}\right\rfloor}+1$ are both in $A$.
1. We start by considering the set \( A \) such that for every \( n \in A \), both \( n^2 + 4 \) and \( \left\lfloor \sqrt{n} \right\rfloor + 1 \) are also in \( A \).

2. Let's analyze the condition \( n \in A \Rightarrow n^2 + 4 \in A \). This implies that if \( n \) is in \( A \), then \( n^2 + 4 \) must also be in \( A \).

3. Next, consider the condition \( n \in A \Rightarrow \left\lfloor \sqrt{n} \right\rfloor + 1 \in A \). For \( n > 1 \), we need to evaluate \( \left\lfloor \sqrt{n^2 + 4} \right\rfloor + 1 \).

4. Notice that for \( n > 1 \):
   \[
   n^2 < n^2 + 4 < (n+1)^2
   \]
   Taking the square root of the inequality:
   \[
   n < \sqrt{n^2 + 4} < n + 1
   \]
   Therefore:
   \[
   \left\lfloor \sqrt{n^2 + 4} \right\rfloor = n
   \]
   Hence:
   \[
   \left\lfloor \sqrt{n^2 + 4} \right\rfloor + 1 = n + 1
   \]

5. This implies that if \( n \in A \), then \( n + 1 \in A \). Therefore, if \( n \in A \), then \( n, n+1, n+2, \ldots \in A \).

6. Now, we need to determine the smallest possible value of \( a \) such that \( \left\lfloor \sqrt{a} \right\rfloor + 1 \geq a \). This condition ensures that the sequence starts correctly.

7. For \( a = 2 \):
   \[
   \left\lfloor \sqrt{2} \right\rfloor + 1 = 1 + 1 = 2
   \]
   This satisfies the condition \( \left\lfloor \sqrt{a} \right\rfloor + 1 \geq a \).

8. Therefore, the smallest starting value \( a \) is 2, and the set \( A \) must be \( \{2, 3, 4, 5, \ldots\} \).

9. We conclude that the only nonempty subset \( A \) that satisfies the given conditions is \( \{2, 3, 4, 5, \ldots\} \).

The final answer is \( \boxed{ \{2, 3, 4, 5, \ldots\} } \).