If $\widehat f(0)=|\widehat f(y)|$ for some $y\neq 0$, let $\theta$ such that $|\widehat f(y)|=e^{i\theta}\widehat f(y)$. Then $$\int_{\Bbb R}f(t)dt=\int_{\Bbb R}f(t)e^{i(ty+\theta})dt,$$ which gives $$\int_{\Bbb R}f(t)(1-e^{i(ty+\theta)})dt=0=\int_{\Bbb R}f(t)(1-\cos(ty+\theta))-i\int_{\Bbb R}f(t)\sin(ty+\theta)dt.$$ In particular $$\int_{\Bbb R}\underbrace{f(t)(1-\cos(ty+\theta))}_{\geq 0}dt=0,$$ so $f(t)(1-\cos(ty+\theta))=0$ almost everywhere. As $f>0$ almost everywhere, we get a contradiction.