Example 1 If the equation $2^{2 x}+2^{x} a+a+1=0$ has real roots for $x$, then the range of values for the real number $a$ is
Analyzing the problem, it seems quite challenging. Instead of tackling it head-on, let's first consider a simpler, similar problem to lay the groundwork. Once we have a clearer path, we can then proceed to solve and conquer the original problem. Let's start with the following lead problem: If the equation \( a = 2^x \) has real roots, then the range of the real number \( a \) is \(\qquad\).

The lead problem is relatively straightforward. Let's set \( y_1 = a \) and \( y_2 = 2^x \). The original equation has a solution if and only if the two newly constructed functions intersect. From the graph, it is clear that the range of \( a \) is the range of \( y_2 = 2^x \). Therefore, the range of \( a \) is \((0, +\infty)\).

Now, we can solve the original problem.
Separating the variables, we get \( a = \frac{-2^{2x} - 1}{2^x + 1} = -\left(2^x + 1\right) - \frac{2}{2^x + 1} + 2 \leq -2 \sqrt{2} + 2 \), so \( a \leq -2 \sqrt{2} + 2 \).