Question: For which two-variable polynomials $p$ with complex coefficients does the equation $p(z,e^z)=0$ have no roots in $\mathbb C$?
The existence of any non-trivial solutions, ie. those not of the form $Ae^{nz}$ for constant $A\neq0$ and integer $n\geq0$, would already be interesting.
Note that $p(z,e^z)$ is an entire function of order 1, so if it has finitely many roots $\rho_1,\ldots,\rho_n$ then by the Hadamard factorisation theorem we can write $$p(z,e^z)=e^{az+b}(z-\rho_1)\cdots(z-\rho_n).$$ In particular, if $p(z,e^z)$ has no roots then it is the exponential of a linear polynomial.
Now the argument of $p(z,e^z)$ as $z\to\infty$ along the positive real axis must approach a constant. (Note that the argument for an entire function with no roots is globally defined.) But the argument for $e^{az+b}$ grows as $\operatorname{Im}(a)z$, so we see that $a$ must be real. By considering the growth rate of $p(z,e^z)$ it is now easy to show that $a\in\mathbb Z_{\geq0}$, so $p(z,e^z)$ must be of the "trivial" form stated in the question.