From [Faulhaber's formula][1] we can see that $S(x,y)*(y+1)*den(\max\{B_y,B_{y-1}\})$ is divisible by $x$, so for $x>z*(y+1)*den(\max\{B_y,B_{y-1}\})$ our expression $S(x,y)$ gives remainder more than $z$ modulo $x$ provided $z\not =0$. 


[1]: https://en.wikipedia.org/wiki/Faulhaber%27s_formula