For an integer $n\ge 3$ and a permutation $\sigma=(p_{1},p_{2},\cdots ,p_{n})$ of $\{1,2,\cdots , n\}$, we say $p_{l}$ is a $landmark$ point if $2\le l\le n-1$ and $(p_{l-1}-p_{l})(p_{l+1}-p_{l})>0$. For example , for $n=7$,\\
the permutation $(2,7,6,4,5,1,3)$ has four landmark points: $p_{2}=7$, $p_{4}=4$, $p_{5}=5$ and $p_{6}=1$. For a given $n\ge 3$ , let $L(n)$ denote the number of permutations of $\{1,2,\cdots ,n\}$ with exactly one landmark point. Find the maximum $n\ge 3$ for which $L(n)$ is a perfect square.
To solve the problem, we need to determine the number of permutations of $\{1, 2, \cdots, n\}$ with exactly one landmark point and find the maximum $n \geq 3$ for which this number is a perfect square.

1. **Definition of Landmark Point**:
   A point $p_l$ in a permutation $\sigma = (p_1, p_2, \cdots, p_n)$ is a landmark point if $2 \le l \le n-1$ and $(p_{l-1} - p_l)(p_{l+1} - p_l) > 0$. This means $p_l$ is either a local maximum or a local minimum.

2. **Conditions for Landmark Points**:
   - If $p_l$ is a local maximum, then $p_{l-1} < p_l$ and $p_{l+1} < p_l$.
   - If $p_l$ is a local minimum, then $p_{l-1} > p_l$ and $p_{l+1} > p_l$.

3. **Counting Permutations with One Landmark Point**:
   To have exactly one landmark point at position $k$, we must have:
   - All numbers $p_j < p_k$ for $j < k$ and $p_k > p_j$ for $j > k$ (local maximum).
   - All numbers $p_j > p_k$ for $j < k$ and $p_k < p_j$ for $j > k$ (local minimum).

4. **Number of Permutations for Each Case**:
   - For a local maximum at position $k$, $p_k = n$. The remaining $n-1$ elements can be arranged in $2^{n-1} - 2$ ways (each element can go to the left or right of $p_k$, but we exclude the cases where $k = 1$ or $k = n$).
   - For a local minimum at position $k$, $p_k = 1$. Similarly, the remaining $n-1$ elements can be arranged in $2^{n-1} - 2$ ways.

5. **Total Number of Permutations with One Landmark Point**:
   \[
   L(n) = 2 \times (2^{n-1} - 2) = 4(2^{n-2} - 1)
   \]

6. **Finding the Maximum $n$ for which $L(n)$ is a Perfect Square**:
   We need to find the maximum $n \geq 3$ such that $L(n) = 4(2^{n-2} - 1)$ is a perfect square.

   - For $n = 3$:
     \[
     L(3) = 4(2^{3-2} - 1) = 4(2^1 - 1) = 4(2 - 1) = 4 \quad (\text{which is } 2^2)
     \]

   - For $n = 4$:
     \[
     L(4) = 4(2^{4-2} - 1) = 4(2^2 - 1) = 4(4 - 1) = 12 \quad (\text{not a perfect square})
     \]

   - For $n = 5$:
     \[
     L(5) = 4(2^{5-2} - 1) = 4(2^3 - 1) = 4(8 - 1) = 28 \quad (\text{not a perfect square})
     \]

   - For $n = 6$:
     \[
     L(6) = 4(2^{6-2} - 1) = 4(2^4 - 1) = 4(16 - 1) = 60 \quad (\text{not a perfect square})
     \]

   - For $n = 7$:
     \[
     L(7) = 4(2^{7-2} - 1) = 4(2^5 - 1) = 4(32 - 1) = 124 \quad (\text{not a perfect square})
     \]

   Continuing this process, we see that $L(n)$ grows rapidly and does not form a perfect square for $n > 3$.

Conclusion:
The maximum $n \geq 3$ for which $L(n)$ is a perfect square is $n = 3$.

The final answer is $\boxed{3}$