Let $f=f(t), g=g(t) \in k[t]$ be two nonzero polynomials over a field $k$ of characteristic zero. Assume that the following two conditions are satisfied:
(i) $\deg(f)=\deg(g) \geq 1$.
(ii) $k(f,gt)=k(t)$.
Example: $f=g=t^2$, indeed, $\deg(f)=\deg(g)=2$ and $k(t^2,t^3)=k(t)$.
Question: Is it possible to find all such $f,g$?
My partial answer: It is not difficult to see that the following three families of $f,g$ are satisfying conditions (i) and (ii):
(1) $\deg(f)=deg(g)=1$, namely, both are linear. Indeed, write $f=at+b, g=ct+d$. Trivially, $at+b \in k(at+b,(ct+d)t)=k(f,gt)$, then $t \in k(f,gt)$, hence $k(f,gt)=k(t)$.
(2) $f=at^n+b$, $g=ct^n+d$, where $a,c \in k-\{0\}$, $b,d \in k$, $n \geq 2$ (actually, we can take $n=1$ and obtain the previous case (1)). Indeed, $at^n+b \in k(at^n+b,(ct^n+d)t)=k(f,gt)$, so $t^n \in k(f,gt)$, and then $g=ct^n+d \in k(f,gt)$, and finally, $t=\frac{gt}{g} \in k(f,gt)$.
(3) $f=g$; this is clear.
Remarks:
(a) I excluded the case $\deg(f)=\deg(g)=0$, because in this case $f=\lambda \in k-\{0\}$, $g=\mu \in k-\{0\}$ trivially satisfy $k(f,gt)=k(\lambda,\mu t)=k(t)$.
(b) The resultant is relevant, since $\operatorname{Res}(f,gt)=\operatorname{Res}(f,g)\operatorname{Res}(f,t)$. However, I am not assuming that $k$ is algebrically closed, so the known property that the resultant of two polynomials is zero iff they have a common root in that algebraically closed field-- is not valid here.
(c) Actually, the $D$-resultant is more relevant, see Theorem 3.2, which implies that the $D$-resultant of $f$ and $gt$ is nonzero.
(d) This is also relevant, though perhaps not practical.
(e) Denote $\deg(f)=\deg(g)=n$. Notice that if $n \geq 2$, then by Abhyankar-Moh-Suzuki's theorem, necessarily $k[f,gt] \subsetneq k[t]$. Indeed, $n=\deg(f) \nmid \deg(gt)=n+1$ and $n+1=\deg(gt) \nmid \deg(f)=n$.
Any hints and comments are welcome!
If $\deg f, \deg h$ are coprime (which is true in your case, when $h=gt$), then $[k(t):k(f)]=\deg f$ and $[k(t):k(h)]=\deg h$ and thus $[k(t):k(f,h)]$ divides both $\deg f, \deg h$ and thus must be one. So, any choice of $f,g$ with $\deg f=\deg g$ will do.