Solution:

Let the legs of the right triangle be $7x$ and $24x$. The hypotenuse is then $c = \sqrt{(7x)^2 + (24x)^2} = \sqrt{49x^2 + 576x^2} = \sqrt{625x^2} = 25x$.

Let $A$, $B$, $C$ be the vertices of the triangle, with right angle at $A$. Let $AB = 7x$, $AC = 24x$, $BC = 25x$.

The incenter $I$ and circumcenter $O$ of a right triangle are separated by a distance $OI = \sqrt{R(R - 2r)}$, where $R$ is the circumradius and $r$ is the inradius.

First, compute $R$ and $r$:

- The circumradius $R$ of a right triangle is half the hypotenuse: $R = \dfrac{25x}{2}$.
- The inradius $r$ is $r = \dfrac{a + b - c}{2}$, where $a$ and $b$ are the legs, $c$ is the hypotenuse.

So,
$$r = \frac{7x + 24x - 25x}{2} = \frac{6x}{2} = 3x.$$

Given $OI = 1$:
$$OI = \sqrt{R(R - 2r)} = 1$$
Plug in $R$ and $r$:
$$\sqrt{\frac{25x}{2}\left(\frac{25x}{2} - 2 \cdot 3x\right)} = 1$$
$$\sqrt{\frac{25x}{2}\left(\frac{25x}{2} - 6x\right)} = 1$$
$$\frac{25x}{2} - 6x = \frac{25x - 12x}{2} = \frac{13x}{2}$$
So,
$$\sqrt{\frac{25x}{2} \cdot \frac{13x}{2}} = 1$$
$$\sqrt{\frac{325x^2}{4}} = 1$$
$$\frac{\sqrt{325}x}{2} = 1$$
$$x = \frac{2}{\sqrt{325}} = \frac{2}{5\sqrt{13}}$$

The area $A$ of the triangle is:
$$A = \frac{1}{2} \cdot 7x \cdot 24x = \frac{1}{2} \cdot 168x^2 = 84x^2$$
Plug in $x$:
$$A = 84 \left(\frac{2}{\sqrt{325}}\right)^2 = 84 \cdot \frac{4}{325} = \frac{336}{325}$$

Final Answer:

The area of the triangle is $\boxed{\dfrac{336}{325}}$.