Is $x^2-ax+a^x-1$ real and less than $2$?
I've begun with finding the discriminant of the function as for a function to be real it's domain needs to be real, but there I'm stuck with another quadratic equation, $a^2-4a^x-4$ how should I proceed to solve this.
As @Robert Shore commented, consider the function $$f(x)=x^2-ax+a^x-3$$ Since there is no restriction on $x$, we must consider $a>0$.
For this function $$f'(x)=2x-a+a^x\log(a) \qquad \text{and} \qquad f''(x)=2+a^x\log^2(a)$$
The first derivative cancels at $$x_*=\frac{a}{2}-\frac{W\left(\frac{1}{2} a^{a/2} \log ^2(a)\right)}{\log (a)}> 0 \quad \forall a >0$$ where $W(\cdot)$ is Lambert funtion.
$$f(x_*)=\frac{W\left(\frac{1}{2} a^{a/2} \log ^2(a)\right) \left(W\left(\frac{1}{2} a^{a/2} \log ^2(a)\right)+2\right)}{\log ^2(a)}-\frac{a^2}{4}-3< 0 \quad \forall a >0$$ $$f''(x_*)=2 \left(W\left(\frac{1}{2} a^{a/2} \log ^2(a)\right)+1\right)> 0 \quad \forall a >0$$
So, $x_*$ corresponds to a minimum of the function.
So $f(x) >0$ as soon as $x$ is greater than the solution of $f(x)=0$. The problem is that this solution does not show analytical expression even using special functions. This would require numerical methods.
Edit
For the solution of $f(x)=0$, we can try approximations using Taylor expansion around $x=x_*$. This gives $$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2+O((x-x_*)^3)$$ and then the estimate of the two roots $$x_\pm=x_* \pm \sqrt{-2 \frac{f(x_*)}{f''(x_*)}}$$ which is not too bad as shown in the table below $$\left( \begin{array}{ccccc} a & x_-^{est} & x_-^{sol} & x_+^{est} & x_+^{sol} \\ 0.5 & -0.903913 & -0.864409 & 1.89548 & 1.92296 \\ 1.0 & -1.000000 & -1.000000 & 2.00000 & 2.00000 \\ 1.5 & -0.936168 & -0.947495 & 1.93926 & 1.92477 \\ 2.0 & -0.812208 & -0.856835 & 1.82693 & 1.76912 \\ 2.5 & -0.687117 & -0.766724 & 1.71781 & 1.62278 \\ 3.0 & -0.578848 & -0.686196 & 1.62603 & 1.50937 \\ 3.5 & -0.489248 & -0.616558 & 1.55184 & 1.42444 \\ 4.0 & -0.415757 & -0.556942 & 1.49223 & 1.36021 \\ 4.5 & -0.355212 & -0.505957 & 1.44407 & 1.31065 \\ 5.0 & -0.304866 & -0.462221 & 1.40477 & 1.27158 \end{array} \right)$$
Using these estimates as $x_0$, Newton method should converge quite fast using, as usual, $$x_{n+1}=x_n-\frac {f(x_n)}{f'(x_n)}$$ In the present case $$x_{n+1}=\frac{a^{x_n} (x_n \log (a)-1)+x_n^2+3}{a^{x_n} \log (a)-a+2 x_n}$$ Using it for a bad case $a=5$, the iterates would be
$$\left( \begin{array}{cc} n & x_n \\ 0 & -0.30486609 \\ 1 & -0.47148352 \\ 2 & -0.46224714 \\ 3 & -0.46222053 \end{array} \right)$$
$$\left( \begin{array}{cc} n & x_n \\ 0 & 1.40477251 \\ 1 & 1.28842795 \\ 2 & 1.27188642 \\ 3 & 1.27157994 \\ 4 & 1.27157984 \end{array} \right)$$