So here, I have the following $$ f(x) = 4^x + 2^x + 1 $$
Now I tried to find the minimum value by doing $y = f(x)$, then doing $4^x + 2^x + 1 - y = 0$, which I solve with the $D \geq 0 $, since this is a real quadratic.
This gives me $y \in [{\frac{3}{4}}, \infty)$, which is not correct here, since $\frac{dy}{dx} = 0$ says me that when $x \rightarrow -\infty$, $y \rightarrow 1$.
Could anyone please explain me why doing with the quadratic way is wrong here, since notice that this is a quadratic of $2^x$?
Your idea of thinking of $y$ as $t^2+t+1$ is often a really useful technique.
In this case we can write $t^2+t+1=(t+\frac{1}{2})^2+\frac{3}{4}$ which you then correctly state has a minimum of $\frac{3}{4}$.
However, this minimum is attained for $t=-\frac{1}{2}$ and this is not, of course, a possible value for $t=e^x$.
I hope this clarifies why your two methods give apparently 'different' answers.