Prove that $\sin x+\sin y=1$ does not have integer solutions

No, and there is not even a solution for $(x,y)\in\mathbb Q\times \mathbb Q$.

We can quickly exclude $x=y$, which would require that $\sin x=\frac12$, but that is only true for $x=n\frac{\pi}{6}$ for certain nonzero integers $n$, and none of these produce a rational. Similarly we can easily exclude $x=0$, $y=0$, or $x=-y$.

Now, using Euler's formula, rewrite the equation to $$ \tag{*} e^{ix} + e^{iy} - e^{-ix} - e^{-iy} = 2i\cdot e^0 $$ and apply the Lindemann–Weierstrass theorem which in one formulation says that the exponentials of distinct algebraic numbers are linearly independent over the algebraic numbers. But $\{\pm ix,\pm iy,0\}$ are all algebraic and (by our assumptions so far) different, so $\text{(*)}$ would be one of the linear relations that can't exist.

This argument generalizes to show that the only algebraic number that can be written as a rational combination of sines of algebraic (radian) angles is $0$.