FindIntegerNullVector is your friend. Notice that this method allows you to obtain an even simpler expression.
A. Find relations between sets of different functions
v[1]={ArcCsc[18/Sqrt[50+17 Sqrt[2]]],ArcTan[Sqrt[2]],ArcCot[3+Sqrt[2]]};
v[2]={ArcCot[Sqrt[2]],ArcCot[2 Sqrt[2]],ArcCot[3+Sqrt[2]]};
v[3]={ArcCot[2 Sqrt[2]],ArcCot[Sqrt[2]],ArcTan[Sqrt[2]]};
v[4]={ArcCot[Sqrt[2]],ArcTan[Sqrt[2]],π/2};
B. This is the main part
Table[FindIntegerNullVector[N[v[i],1000]],{i,4}]
Out[1]= {{-3,-1,11},{-2,1,4},{1,1,-1},{1,1,-1}}
C. Make a list of replacement rules
r[1]={ArcCsc[18/Sqrt[50+17 Sqrt[2]]]->(-ArcTan[Sqrt[2]]+11ArcCot[3+Sqrt[2]])/3};
r[2]={ArcCot[3+Sqrt[2]]->1/4(2ArcCot[Sqrt[2]]-ArcCot[2 Sqrt[2]])};
r[3]={ArcCot[2 Sqrt[2]]->ArcTan[Sqrt[2]]-ArcCot[Sqrt[2]]};
r[4]={ArcCot[Sqrt[2]]->π/2-ArcTan[Sqrt[2]]};
D. Apply the rules
sx=Fold[ReplaceAll,x,Array[r,4]]//Expand;
sy=Fold[ReplaceAll,y,Array[r,4]]//Expand;
sx-sy
Out[2]= 0
Comment/observation
In principle, all but one identities found by the LLLLLL method are easy to prove by hand. Only the first one, namely $$11\,\mathrm{arccot}\big(3+\sqrt{2}\big) = \mathrm{arctan}\big(\sqrt{2}\big)+ 3\,\mathrm{arccsc}\!\Bigg\{\frac{18}{\sqrt{50+17 \sqrt{2}}}\Bigg\}$$ is somewhat unexpected.