Let $x = \tan \alpha$, $y = \tan \beta$.

We have $\cos^2 \alpha = \dfrac{1}{1 + x^2}$ and $\cos^2 \beta = \dfrac{1}{1 + y^2}$.

So,
$$\left( \dfrac{1}{1 + x^2} + \dfrac{1}{1 + y^2} \right)(1 + x y) = 2.$$

Multiply both sides by $(1 + x^2)(1 + y^2)$:
$$\left[ (1 + y^2) + (1 + x^2) \right](1 + x y) = 2(1 + x^2)(1 + y^2)$$
$$(2 + x^2 + y^2)(1 + x y) = 2(1 + x^2 + y^2 + x^2 y^2)$$

Expand the left side:
$$(2 + x^2 + y^2)(1 + x y) = 2(1 + x y) + x^2(1 + x y) + y^2(1 + x y)$$
$$= 2(1 + x y) + x^2 + x^3 y + y^2 + x y^3$$

But let's expand both sides fully:

Left:
$$(2 + x^2 + y^2)(1 + x y) = 2(1 + x y) + x^2(1 + x y) + y^2(1 + x y)$$
$$= 2 + 2 x y + x^2 + x^3 y + y^2 + x y^3$$

Right:
$$2(1 + x^2)(1 + y^2) = 2(1 + x^2 + y^2 + x^2 y^2) = 2 + 2x^2 + 2y^2 + 2x^2 y^2$$

Set equal:
$$2 + 2 x y + x^2 + x^3 y + y^2 + x y^3 = 2 + 2x^2 + 2y^2 + 2x^2 y^2$$

Subtract $2$ from both sides:
$$2 x y + x^2 + x^3 y + y^2 + x y^3 = 2x^2 + 2y^2 + 2x^2 y^2$$

Bring all terms to one side:
$$2 x y + x^2 + x^3 y + y^2 + x y^3 - 2x^2 - 2y^2 - 2x^2 y^2 = 0$$
$$2 x y + x^2 + x^3 y + y^2 + x y^3 - 2x^2 - 2y^2 - 2x^2 y^2 = 0$$
Group terms:
$$(x^2 - 2x^2) + (y^2 - 2y^2) + 2 x y + x^3 y + x y^3 - 2x^2 y^2 = 0$$
$$(-x^2) + (-y^2) + 2 x y + x^3 y + x y^3 - 2x^2 y^2 = 0$$

Now, $x^3 y + x y^3 = x y (x^2 + y^2)$, so:
$$(-x^2) + (-y^2) + 2 x y + x y (x^2 + y^2) - 2x^2 y^2 = 0$$
$$(-x^2 - y^2) + 2 x y + x y (x^2 + y^2) - 2x^2 y^2 = 0$$

Now, $x y (x^2 + y^2) = x^3 y + x y^3$ as above.

Group $-x^2 - y^2$ and $x y (x^2 + y^2)$:
$$2 x y + x y (x^2 + y^2) - x^2 - y^2 - 2x^2 y^2 = 0$$

Let us factor $x y$:
$$2 x y + x y (x^2 + y^2) = x y [2 + x^2 + y^2]$$
So:
$$x y [2 + x^2 + y^2] - x^2 - y^2 - 2x^2 y^2 = 0$$

Bring $x^2 + y^2$ to the other side:
$$x y [2 + x^2 + y^2] = x^2 + y^2 + 2x^2 y^2$$

Now, move all terms to one side:
$$x y [2 + x^2 + y^2] - x^2 - y^2 - 2x^2 y^2 = 0$$

Alternatively, let's try the hint: write the given equality as $(\tan \alpha \tan \beta - 1)(\tan \alpha - \tan \beta)^2 = 0$.

Let us try to factor the expression.

Let $x = \tan \alpha$, $y = \tan \beta$.

Suppose $x y - 1 = 0$, i.e. $x y = 1$.

Then $\tan \alpha \tan \beta = 1$.

Recall that $\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.

If $\tan \alpha \tan \beta = 1$, then denominator is $0$, so $\tan (\alpha + \beta)$ is undefined, which happens when $\alpha + \beta = 90^\circ$.

Since $\alpha$ and $\beta$ are acute and $\alpha \neq \beta$, this is possible.

Alternatively, $(x - y)^2 = 0$ gives $x = y$, i.e. $\tan \alpha = \tan \beta$, so $\alpha = \beta$, but $\alpha \neq \beta$ by assumption.

Therefore, the only possibility is $\tan \alpha \tan \beta = 1$, i.e. $\alpha + \beta = 90^\circ$.

Thus, $\boxed{\alpha + \beta = 90^\circ}$.