Two numbers that cannot both be squares
Solution 1:
Consider $a < b$:
Clearly, $b^2 < a+b^2$. Further, we see that $$ a+b^2 < b + b^2 = b(b+1) < (b+1)^2 $$ Hence, $b^2 < a+b^2 < (b+1)^2$. Thus, $a+b^2$ is not a square. On the other hand, $a^2+b$ may be a square, depending on the choice of $a$ and $b$.
If we have $a=b$, then neither $a^2+b$ nor $a+b^2$ are squares, as they would both simplify to $a(a+1)$.