Partly logic, partly not

$\begin{aligned} \color{#D61F06} y&\color{#D61F06} >x \\ \color{#3D99F6} y&\color{#3D99F6} >x+1 \\ \color{#20A900} y&\color{#20A900}>x+2 \\ \color{#D61F06} y-x&\color{#D61F06}>0 \\ \color{#3D99F6} y-x&\color{#3D99F6}>1 \\ \color{#20A900} y-x&\color{#20A900}>2 \\ \color{#69047E} x+y&\color{#69047E}=99 \end{aligned}$

Note that if two inequalities are colored with the same color, then if any of them is true, then the other is also true and if any of them is false, then the other is also false.

If there are $n$ number of true inequalities out of the first six inequalities, then $n$ is even, and $3$ is odd, so $x+y=99$ has to be true. From the first six inequalities exactly two is true.

Note that if $y>x+2$ , then $y>x+1$ , furthermore if $y>x+1$ , then $y>x$ . The same can be said about the the 4th, 5th, 6th inequalities, so it is clear that the two ${\color{#D61F06}{\text{red inequalities}}}$ and the ${\color{#69047E}{\text{equation}}}$ are the three true statements.

Since $x+y=99$ , and out of the $x, y$ numbers, at least one is an integer, the other is an integer too. We know that $\begin{aligned} y & > x \\ y & \not > x+1\end{aligned}$ so the only solution is: $\begin{aligned} x & =49 \\ y & =50 \end{aligned}$

Therefore the answer is $x^2-y=49^2-50=\boxed{2351}$