To find all pairs of positive integers $(x, y)$ such that $(xy+1)(xy+x+2)$ is a perfect square, we start by analyzing the given expression.

First, consider the greatest common divisor (gcd) of the two terms:
$$\gcd(xy+1, xy+x+2).$$
Using the properties of gcd, we have:
$$\gcd(xy+1, xy+x+2) = \gcd(xy+1, x+1).$$
This simplifies further to:
$$\gcd(xy+1, x+1) = \gcd(y-1, x+1).$$

Let $x+1 = da$ and $y-1 = db$ where $\gcd(a, b) = 1$. Then we can express $xy+1$ and $xy+x+2$ as:
$$xy+1 = d \cdot u^2 \quad \text{and} \quad xy+x+2 = d \cdot v^2,$$
for some relatively prime integers $u$ and $v$.

Using the relationship $a = v^2 - u^2$, we can rewrite the equation as:
$$u^2 = (d \cdot b + 1)(v^2 - u^2) - b,$$
or equivalently:
$$(d \cdot b + 1)v^2 - (d \cdot b + 2)u^2 = b.$$

This is a form of a Pell equation. To solve this, note that $v > u$. Let $v = \frac{X+Y}{2}$ and $u = \frac{X-Y}{2}$ for positive integers $X$ and $Y$. Substituting these into the equation, we get:
$$X^2 - (4bd + 6)XY + Y^2 + 4b = 0.$$

Using Vieta jumping, assume there is a solution $(X, Y)$ in positive integers with $X \ge Y$. By symmetry, the pair $\left( \frac{Y^2+4b}{X}, Y \right)$ is also a solution. Repeating this process, we eventually reach pairs $(X_1, Y)$ and $(X_2, Y)$ with $X_1 > X_2 \ge Y$. This implies:
$$\begin{align*} X_1 + X_2 &= (4bd + 6)Y, \\ X_1 \cdot X_2 &= Y^2 + 4b. \end{align*}$$

If $\min(X_1, X_2) > Y$ and $X_1 + X_2 = (4bd + 6)Y$, then:
$$X_1 \cdot X_2 \ge Y \cdot (4bd + 5)Y > Y^2 + 4b,$$
which leads to a contradiction.

Thus, there are no pairs of positive integers $(x, y)$ such that $(xy+1)(xy+x+2)$ is a perfect square.

The answer is: \boxed{\text{No solutions}}.