How to calculate expected payoff and optimal bid in this auction?

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There is a seller $S$ with an object with quality $x$. The seller knows this quality; the buyer $B$ does not but beliefs that $x$ is uniformly distributed on $[0,M]$ with $0<M\leq2$.

With true quality $x$, the seller values the object with $x$ and the buyer with $x^2$.

The game is:

  1. the seller learns $x$,

  2. the buyer makes a bid $b$,

  3. the seller accepts or rejects,

  4. the payoff to the seller is $b-x$ and the payoff to the buyer is $x^2-b$. If the bid is rejected both get $0$. For each value of $M$, find a PBE of this game.

What I have done so far:

Note that for uniform distribution on $[0,M]$ the pdf is $f(x) = \frac{1}{M}$ and the cdf is $F(x) = \frac{x}{M}$ for $x \in [0,M]$.

Clearly, for the seller the optimal strategy is to accept bids when $b>x$ and reject bids $b<x$. The expected payoff of the buyer is then: $$Eu_B = \int_0^b x^2 - b\ \textrm{d}F(x)$$ Where $F(x)$ is cdf of the uniform distribution. Correct?

From this answer I understand that this is equivalent to: $$Eu_B = \int_0^b (x^2 - b)f(x)\ \textrm{d}x$$

Substituting the pdf the gives: $$Eu_B = \int_0^b \frac{(x^2 - b)}{M}\ \textrm{d}x = \frac{\frac{1}{3} b^3 - b^2}{M}$$

Setting the derivative w.r.t. $b$ to zero: $$\frac{\partial Eu_B}{\partial b} = \frac{b^2 - 2b}{M} = 0 \iff b = 2$$

But this doesn't make sense to me. Why would the buyer always bid 2, regardless of $M$? What am I missing here?

EDIT: I am not looking for a complete solution here, just some pointers in the right direction, or correction of any mistakes in my solution.