The functions $\{ \sin(nx) \}_{n=1}^{\infty}$ form a complete orthogonal basis for $L^2(0,\pi)$ because they are the eigenfunction solutions of
$$
-y'' = \lambda y,\;\;\; y(0)=y(\pi)=0.
$$
The normalization constants are
$$
\int_{0}^{\pi}\sin^2(nx)dx=\frac{1}{2}\int_{0}^{2\pi}\sin^2(nx)dx
= \frac{1}{4}\int_{0}^{2\pi}\sin^2(nx)+\cos^2(nx)dx = \frac{\pi}{2}.
$$
Any $f\in L^2(0,\pi)$ can be written uniquely as $\sum_{n=1}^{\infty}f_n\sin(nx)$ where $\{ f_n \}_{n=1}^{\infty} \in \ell^2$. In fact,
$$
f_n = \frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx,\;\; n=1,2,3,\cdots.
$$
So your problem can be reduced to a problem on $\ell^2(\mathbb{N})$, where $u$ is represented by $\{ u_n \}_{n=1}^{\infty}$. The eigenfunction problem becomes a coefficient problem in $\ell^2$:
$$
Ku=\lambda u \\
\int_{0}^{\pi}u(y)k(x,y)dy=\lambda u(x) \\
\int_{0}^{\pi}u(y)\sum_{n=1}^{\infty}\frac{1}{n^2}\sin((n+1)x)\sin(ny)dy=\lambda u(x)
$$
This gives coefficient equations after rewriting as
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n^2}\sin((n+1)x)\frac{\pi}{2}u_n&=\sum_{n=1}^{\infty}\lambda u_n\sin(nx) \\
\sum_{n=2}^{\infty}\frac{1}{(n-1)^2}\sin(nx)\frac{\pi}{2}u_{n-1} & = \sum_{n=1}^\infty \lambda u_n \sin(nx).
\end{align}
There is no $\sin(x)$ term on the left, which forces $\lambda u_1=0$. The general $u_n$ must satisfy
$$
\frac{\pi}{2}\frac{1}{(n-1)^2}u_{n-1}=\lambda u_n,\;\; n \ge 2.
$$
If $\lambda \ne 0$, then $u_1 =0$ and every $u_n=0$ for $n > 1$ by the above. If $\lambda = 0$, then every $u_n=0$ by the above. So there are no eigenvalues of this operataor. That means that the operator $K$ is quasinilpotent with spectrum $\{0\}$.
