Note that the sine function is periodically and $$\sqrt{n^2 + n} - n =\frac{n}{\sqrt{n^2 +n} + n}$$

Therefore,

$$\lim_{n\to\infty}\sin^2\left(\pi\sqrt{n^2+n}\right)=\lim_{n\to\infty}\sin\left(\pi\sqrt{n^2+n}-n\pi\right)=\lim_{n\to\infty}\sin^2\left(\frac{n\pi}{\sqrt{n^2+n}+n}\right)=\boxed{1}$$