Proof: From the condition, we know
$$b^{2}=2 a^{2}+a-\left(2 b^{2}+b\right)=(a-b)(2 a+2 b+1) .$$

The left side of the above equation is greater than zero, and on the right side, $2 a+2 b+1$ is greater than zero, so $a-b$ is greater than zero.
From (1), to prove that $a-b$ and $2 a+2 b+1$ are both perfect squares, it is only necessary to prove
$$(a-b, 2 a+2 b+1)=1$$

Let $(a-b, 2 a+2 b+1)=d$, then from (1) we know $d^{2} \mid b^{2}$, so $d \mid b$. Furthermore, combining $d \mid a-b$, we know $d \mid a$, hence $d \mid 2(a+b)$. Since $d \mid 2 a+2 b+1$, it follows that $d \mid 1$, thus $d=1$.

The proposition is proved.