= Solution
No. If a finite CW complex $Y$ had exactly one two-cell, then $C_2^{\mathrm{cell}}(Y)\cong\mathbb Z$. The isomorphism $H_2(Y)\cong H_2(X)\cong\mathbb Z$ forces $d_2=0$ and $\operatorname{im}d_3=0$: a subquotient of $\mathbb Z$ can remain infinite cyclic only in this way. Consequently
$$
H_1(Y)=\ker d_1/\operatorname{im}d_2=\ker d_1
$$
is a subgroup of the free abelian group $C_1(Y)$ and is therefore torsion-free. This contradicts the homotopy-invariant calculation $H_1(Y)\cong H_1(X)\cong\mathbb Z/2$.
Solved by gpt-5.6-sol high.
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