Not a proof, but some considerations which could be developed into a proof.
Let's change the variable to $$x=\sinh t$$
Now the function becomes:
$$f(t)=e^{\sinh t} \cos (\cosh t), \quad t \in [0, \ln (1+\sqrt{2})]$$
Note that $$\ln (1+\sqrt{2}) < \ln 2.5 < 1$$
This function is easy to differentiate:
$$f'(t)=e^{\sinh t} (\cosh t \cos (\cosh t) - \sinh t \sin (\cosh t))$$
The extrema obey the equation:
$$\tanh t \tan (\cosh t)=1$$
Remember that: $$\cosh t \geq 1 > \frac{\pi}{4}$$
Considering the ranges of both tangents, it's not hard to guess that there should only be a single solution in our interval.
To approximate the value, we can use bisection method:
$$g(t)=\tanh t \tan (\cosh t)-1$$
$$g(0) = -1 <0 \\ g(\ln (1+\sqrt{2})=\frac{\tan \sqrt{2}}{\sqrt{2}}-1 >0$$
Because for the range $a \in [0, \pi/2]$ we have $\tan a > a$.
Now we check the middle of the interval:
$$g(1/2)=\tanh \frac{1}{2} \tan \left(\cosh \frac{1}{2} \right)-1=\frac{e-1}{e+1} \tan \frac{e+1}{2 \sqrt{e}}-1=-0.0264\dots$$
We had to use calculators here, but with some care and knowledge of the first few digits of $e$ we can at least prove the the value is negative.
Which means that the extremum point $t_0 \in (1/2,1)$. However, we don't need to check the middle of this new interval, because $g(1/2)$ is much smaller than $g(0)$ and it's quite clear that $t_0$ is close to $1/2$. Moreover, we know* that the function $f(t)$ is growing on the interval $(0,1/2)$. So we can write:
$$f(t)< f(1/2) = e^{\sinh \frac{1}{2}} \cos \left( \cosh \frac{1}{2} \right)=0.722 \dots$$
This is again done with a calculator, however we can fix that by picking some number close to $1/2$ which gives us a more simple expression.
$^*$ It remains to be shown that $t_0$ is a maximum, but it can be done in principle by checking a few suitable values of $f(t)$ around it.