Let lengths of three sides be $z -1$, $z$, and $z + 1$, respectively.

Then the semi-perimeter is $\frac{3}{2}z$ and the area is $$S=\sqrt{\frac{3}{2}z\cdot\Big(\frac{3}{2}z-(z-1)\Big)\Big(\frac{3}{2}z-z\Big)\Big(\frac{3}{2}z-(z+1)\Big)}=\frac{z}{4}\sqrt{3\cdot(z^2-4)}$$

If $S$ is an integer, then $3(z^2-4)$ must be a square number. Let $(z^2-4) = 3w^2$.

Furthermore, $z$ must be even because otherwise both $z$ and $\sqrt{z^2-4}$ will be odd. This means that $z\sqrt{3\cdot(z^2-4)}$ cannot be a multiple of $4$.

Let $z=2x$. Hence we have $4x^2 -4 =3w^2$. This means $w$ must be even too. Letting $w=2y$ leads to: $$x^2-3y^2 = 1$$

This is a Pell's equation, with the following general solution: $$ \left\{\begin{array}{ll} x_n = \displaystyle\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2}\\ y_n = \displaystyle\frac{(2+\sqrt{3})^n-(2-\sqrt{3})^n}{2\sqrt{3}} \end{array}\right. $$ Accordingly, the three sides are $2x_n -1$, $2x_n$, and $2x_n+1$. The area is $3 x_n y_n$.