jmc

Let $y=x-1.$ Then $x=y+1,$ and $$(y+1{)}^4+(-y+1{)}^4=34.$$Expanding, we get $2y^4+12y^2-32=0.$ This factors as $2(y^2-2)(y^2+8)=0,$ so $y=\pm \sqrt{2}.$ Thus, the solutions in $x$ are $\boxed{1+\sqrt{2},1-\sqrt{2}}.$