From the relation $Sq(v)=w$ we get that $v_2=w_2+w_1^2$. Now we have a nice characterisation of the (second) Wu-class, as $$Sq^{2}(x)=v_2\smile x$$ for any $x \in H^{n-k}(M;\Bbb Z_2)$ (See Milnor Stasheff page 132). Since in our case $n=3$, we get that for any $x\in H^1(M;\Bbb Z_2)$, $0=v_2\smile x$, which gives you $v_2=0$.
Therefore you have $0=w_2+w_1^2$, which proves your claim