Find all positive integers $x$ such that $2x+1$ is a perfect square but none of the integers $2x+2, 2x+3, \ldots, 3x+2$ are perfect squares.
1. We start by noting that \(2x + 1\) is a perfect square. Let \(2x + 1 = k^2\) for some integer \(k\). Therefore, we have:
   \[
   2x + 1 = k^2 \implies 2x = k^2 - 1 \implies x = \frac{k^2 - 1}{2}
   \]
   Since \(x\) is a positive integer, \(k^2 - 1\) must be even, which implies \(k\) must be odd.

2. Next, we need to ensure that none of the integers \(2x + 2, 2x + 3, \ldots, 3x + 2\) are perfect squares. The next perfect square after \(2x + 1 = k^2\) is \((k + 1)^2\). We need:
   \[
   (k + 1)^2 > 3x + 2
   \]
   Substituting \(x = \frac{k^2 - 1}{2}\) into the inequality, we get:
   \[
   (k + 1)^2 > 3\left(\frac{k^2 - 1}{2}\right) + 2
   \]
   Simplifying the right-hand side:
   \[
   (k + 1)^2 > \frac{3k^2 - 3}{2} + 2 \implies (k + 1)^2 > \frac{3k^2 - 3 + 4}{2} \implies (k + 1)^2 > \frac{3k^2 + 1}{2}
   \]

3. Expanding and simplifying the left-hand side:
   \[
   (k + 1)^2 = k^2 + 2k + 1
   \]
   Therefore, the inequality becomes:
   \[
   k^2 + 2k + 1 > \frac{3k^2 + 1}{2}
   \]
   Multiplying both sides by 2 to clear the fraction:
   \[
   2(k^2 + 2k + 1) > 3k^2 + 1 \implies 2k^2 + 4k + 2 > 3k^2 + 1
   \]
   Simplifying:
   \[
   2k^2 + 4k + 2 > 3k^2 + 1 \implies 4k + 2 > k^2 + 1 \implies k^2 - 4k - 1 < 0
   \]

4. Solving the quadratic inequality \(k^2 - 4k - 1 < 0\) using the quadratic formula:
   \[
   k = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}
   \]
   Since \(k\) must be an integer, we consider the interval \(2 - \sqrt{5} < k < 2 + \sqrt{5}\). Approximating \(\sqrt{5} \approx 2.236\), we get:
   \[
   2 - 2.236 < k < 2 + 2.236 \implies -0.236 < k < 4.236
   \]
   Since \(k\) is an odd integer, the possible values for \(k\) are 1 and 3.

5. Checking these values:
   - For \(k = 1\):
     \[
     2x + 1 = 1^2 \implies 2x + 1 = 1 \implies x = 0 \quad (\text{not a positive integer})
     \]
   - For \(k = 3\):
     \[
     2x + 1 = 3^2 \implies 2x + 1 = 9 \implies 2x = 8 \implies x = 4
     \]

6. Verifying \(x = 4\):
   \[
   2x + 2 = 10, \quad 2x + 3 = 11, \quad \ldots, \quad 3x + 2 = 14
   \]
   None of these values (10, 11, 12, 13, 14) are perfect squares.

Thus, the only positive integer \(x\) that satisfies the given conditions is \(x = 4\).

The final answer is \(\boxed{4}\)