You already solved a major part of the problem. Now just take logarithm of the both sides: $$ \log(F(n)+1) = \log(F(n-1)+1) + \log(F(n-2)+1)$$ Then define $G(n) = \log(F(n)+1)$. Now you have a Fibonacci-like recursive relationship: $$G(n)=G(n-1)+G(n-2)$$ You might know that Fibonacci sequence has a closed form solution in terms of inital terms as follows: $$ U_n=\frac{U_1-U_0\psi}{\sqrt{5}}\times \phi^n+ \frac{U_0\phi-U_1}{\sqrt{5}}\times \psi^n $$ where $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$. Now just replace $U_n$ with $G(n)$, then you have the following: $$ G(n)=\frac{G(1)-G(0)\psi}{\sqrt{5}}\phi^n+ \frac{G(0)\phi-G(1)}{\sqrt{5}}\psi^n $$ Now get back to the $F(n)$ by using $F(n)=e^{G(n)}-1$ which is: $$F(n)=e^{(\frac{\log(F(1)+1)-\log(F(0)+1)\psi}{\sqrt{5}}\phi^n+ \frac{\log(F(0)+1)\phi-G\log(F(1)+1)}{\sqrt{5}}\psi^n)}-1$$

You already solved a major part of the problem. Now just take logarithm of the both sides: $$ \log(F(n)+1) = \log(F(n-1)+1) + \log(F(n-2)+1)$$ Then define $G(n) = \log(F(n)+1)$. Now you have a Fibonacci-like recursive relationship: $$G(n)=G(n-1)+G(n-2)$$ You might know that Fibonacci sequence has a closed form solution in terms of inital terms as follows: $$ U_n=\frac{U_1-U_0\psi}{\sqrt{5}}\times \phi^n+ \frac{U_0\phi-U_1}{\sqrt{5}}\times \psi^n $$ where $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$. Now just replace $U_n$ with $G(n)$, then you have the following: $$ G(n)=\frac{G(1)-G(0)\psi}{\sqrt{5}}\phi^n+ \frac{G(0)\phi-G(1)}{\sqrt{5}}\psi^n $$ Now get back to the $F(n)$ by using $F(n)=e^{G(n)}-1$ which is: $$F(n)=e^{(\frac{\log(F(1)+1)-\log(F(0)+1)\psi}{\sqrt{5}}\phi^n+ \frac{\log(F(0)+1)\phi-\log(F(1)+1)}{\sqrt{5}}\psi^n)}-1$$