What $a$ makes $\{x\mid a^x = x\}$ a singleton?
$$(1.4444)^x - x \le 0 \tag 1$$ has real solutions.
$$(1.4447)^x - x \le 0 \tag 2$$ has no real solutions.
I guess $1.4444 < a < 1.4447$
I tried running simulations using goal seek in Excel, but I think I'm doing it wrong because I keep getting a lot of values below $1.4$.
How can I approach this problem?
Note that $a^x = x \iff a = x^{1/x}$. So, there will be a solution to your problem if and only if $a$ is in the image of the function $f(x) = x^{1/x}$ (over the domain $x > 0$). Note that the graph $y = f(x)$ achieves a maximum somewhere, then levels off to its asymptote at $y = 1$.
This problem is a bit easier to solve with calculus. In particular, it suffices to find the maximum value of $f(x)$.
Note that $f(x)$ achives its maximum iff $\ln(f(x))$ achieves its maximum. So, we consider the function $$ g(x) = \ln(f(x)) = \frac{\ln x}{x} $$ We find $$ g'(x) = \frac{1 - \ln x}{x^2} $$ Thus, $g$ has a unique critical point when $x = e$, which means that this must be where $g$ achieves its maximum. Thus, $f$ achieves its maximum at $x = e$.
Thus, the maximum value of $a$ such that $f(x) = a$ has a solution is $$ a = e^{1/e} \approx 1.44467 $$