Let $F$ be an infinite field and $f$ a homogeneous form on $F$ such that $f$ has no nontrivial zero in $F$. Let $F'$ be a finite extension of $F$ such that $f$ has a nontrivial zero in $F'$. Is it true there exists a simple extension of $F$ of the form $F(\alpha)$ contained in $F'$ which contains a nontrivial zero of $f$?

I believe the following is a negative example: Let $s,t,u,v$ be variables over $\mathbb F_p$. Set $F=\mathbb F_p(s,t,u,v)$ and $F'=F(\sigma,\tau)$ with $\sigma^p=s$, $\tau^p=t$. Set $$f(X,Y,Z)=(X^psZ^p)u+(Y^ptZ^p)v.$$ Then $f(\sigma,\tau,1)=0$. We show that any solution of $f=0$ over $F'$ has this form up to a scalar factor: Let $x,y,z\in F'$ with $f(x,y,z)=0$. As $F'=\mathbb F_p(u,v,\sigma,\tau)$, we get $$x^psz^p, y^ptz^p\in\mathbb F_p(u^p,v^p,\sigma^p,\tau^p)=\mathbb F_p(u^p,v^p,s,t),$$ hence $$A(u^p,v^p,s,t)u+B(u^p,v^p,s,t)v=0$$ for rational functions $A,B$ over $\mathbb F_p$ with $x^psz^p=A(u^p,v^p,s,t)$ and $y^ptz^p=B(u^p,v^p,s,t)$. This implies $A(u^p,v^p,s,t)=0$, for otherwise $u$ were a rational function in $u^p$. For the same reason $B(u^p,v^p,s,t)=0$. We get $x^ptz^p=0=y^ptz^p$, and the claim follows. 

