How many integers $n$ (with $1 \le n \le 2021$) have the property that $8n + 1$ is a perfect square?
1. We start by noting that we need \(8n + 1\) to be a perfect square. Let \(8n + 1 = k^2\) for some integer \(k\). Then we have:
   \[
   k^2 = 8n + 1
   \]
   Rearranging this equation, we get:
   \[
   8n = k^2 - 1 \implies n = \frac{k^2 - 1}{8}
   \]

2. For \(n\) to be an integer, \(\frac{k^2 - 1}{8}\) must also be an integer. This implies that \(k^2 - 1\) must be divisible by 8. We can rewrite \(k^2 - 1\) as:
   \[
   k^2 - 1 = (k - 1)(k + 1)
   \]
   For \((k - 1)(k + 1)\) to be divisible by 8, both \(k - 1\) and \(k + 1\) must be even, and one of them must be divisible by 4. This means \(k\) must be odd.

3. Next, we need to find the range of \(k\) such that \(1 \leq n \leq 2021\). From the equation \(n = \frac{k^2 - 1}{8}\), we have:
   \[
   1 \leq \frac{k^2 - 1}{8} \leq 2021
   \]
   Multiplying through by 8, we get:
   \[
   8 \leq k^2 - 1 \leq 16168
   \]
   Adding 1 to all parts of the inequality, we get:
   \[
   9 \leq k^2 \leq 16169
   \]
   Taking the square root of all parts, we get:
   \[
   3 \leq k \leq \sqrt{16169} \approx 127.16
   \]
   Since \(k\) must be an integer, we have:
   \[
   3 \leq k \leq 127
   \]

4. Since \(k\) must be odd, we list the odd integers between 3 and 127 inclusive. These are:
   \[
   3, 5, 7, \ldots, 125, 127
   \]
   This is an arithmetic sequence with the first term \(a = 3\) and common difference \(d = 2\). The number of terms in this sequence can be found using the formula for the \(n\)-th term of an arithmetic sequence:
   \[
   a_n = a + (n-1)d
   \]
   Setting \(a_n = 127\), we solve for \(n\):
   \[
   127 = 3 + (n-1) \cdot 2
   \]
   \[
   127 = 3 + 2n - 2
   \]
   \[
   127 = 2n + 1
   \]
   \[
   126 = 2n
   \]
   \[
   n = 63
   \]

Therefore, there are 63 integers \(n\) such that \(8n + 1\) is a perfect square.

The final answer is \(\boxed{63}\).