This is a nice one. The idea is to see a bijection between the quadruple $(a,b,x,y)$ to $(x+y,x,b,a-b)$. Notice that one of them works iff the other one does. But there are quadruples that pair up to themselves, and those are such that $a=x+y,b=x$. Substituting this on the equation, we get that $b^2+2by=2023$, which implies $b$ divides 2023 and $b^2<2023$. This implies $b=1,17,7$ which all work. Notice that thus we have an even quantity (because of the bijection) plus 3 cases we checked. Thus $T_{2023}$ is odd, as required.