Answer. There are three such pairs, $(x, y) = (2, 2)$, $(x, y) = (2, 3)$ and $(x, y) = (3, 2)$.

For $x = 1$, we get $d = 1$ and the given equation becomes the contradiction $y = y + 2$. This works analogously for $y = 1$.
Therefore, we can assume $x \ge 2$ and $y \ge 2$.

We start with the case $d = 1$ which gives the equation
$$xy = x + y + 1 \iff (x - 1)(y - 1) = 2.$$
The possible factorizations $2 = 1 \cdot 2$ and $2 = 2 \cdot 1$ give the pairs $(x, y) = (2, 3)$ and $(x, y) = (3, 2)$, respectively, because $\gcd(x, y) = 1$ is satisfied.

Now, we treat the case $d \ge 2$. The given equation is equivalent to
$$\frac{1}{xd} + \frac{1}{yd} + \frac{d}{xy} = 1.$$
Because of $xd \ge 4$ and $yd \ge 4$, we get
$$1 \le \frac{1}{4} + \frac{1}{4} + \frac{d}{xy} \iff xy \le 2d.$$
Together with $xy \ge d^2$, we obtain $d = 2$, $x = y = 2$ which gives indeed the third pair $(x, y) = (2, 2)$ with $\gcd(2, 2) = 2$.