Let´s take any $x>3$ and choose a,b such that $a<b$,$a-b$ even and $ab=x^2-1$.
Then with $y=(b-a)/2$, $z=(b+a)/2$ we have $x^2 + y^2 = z^2 + 1$.
Particular cases∶
if $x$ even then $x^2 + ((x^2-2)/2)^2 = (x^2/2)^2 + 1$; 
if $x$ odd then $x^2 + ((x^2-5)/4)^2 = ((x^2+3)/4)^2 + 1$.