Let the sides be $p, q, r$. By Heron's $A = \sqrt{s(s-p)(s-q)(s-r)}$ where $s = \frac{p+q+r}{2}$. If $p+q+r$ is odd then the area will have some factor of $\frac{1}{4}$ and thus not be an integer. Otherwise, at least one of $p, q, r$ must be $2$. However then by triangle inequality, then we must have $2, p, p$ for prime $p$. Heroning this gives $\sqrt{(p+1)(p-1)}$, and $p^2 - 1$ is never a square for prime $p$.