There are countably infinitely many zeros of your $f(x)$, which we can enumerate as follows.
For small $x$, your function is approximated by $e^{1/x} x \sinh(1/x)$:
f[x_] = Hypergeometric1F1[1 - x, 2, 2/x];
g[x_] = E^(1/x) x Sinh[1/x];
ReImPlot[{f[I x], g[I x]}, {x, -0.1, 0.1}]
This approximation gives excellent starting points for numerical root-finding: for $n\in\mathbb{Z}\setminus\{0\}$ we have $g[i/(n\pi)]=0$, and find the $n^{\text{th}}$ zero of $f$ from this starting point:
zf[n_Integer /; n != 0] := x /. FindRoot[f[x], {x, I/(n*π)}]
Array[zf, 10] // Chop
(* {0. + 0.458929 I, 0. + 0.173708 I,
0. + 0.110616 I, 0. + 0.0815787 I,
0. + 0.0647315 I, 0. + 0.0536934 I,
0. + 0.0458896 I, 0. + 0.0400755 I,
0. + 0.035574 I, 0. + 0.0319845 I} *)
Much higher zeros can be found in this way:
zf[10^6]
(* 0. + 3.1831*10^-7 I *)
For added stability we can extract more factors from $g(x)$ to regularize your $f(x)$ before nulling: notice that $g(i/y) y e^{i y} = \sin y$ with zeros at $y_n=n\pi$, and further that $f(i/y) y e^{i y}\in\mathbb{R}$ if $y\in\mathbb{R}$; and hence
zf2[n_Integer /; n != 0] :=
I/y /. FindRoot[Re[f[I/y] E^(I y) y], {y, n*π}]
zf2[10^6]
(* 0. + 3.1831*10^-7 I *)
ReImPlot[{f[I/y] E^(I y) y, Sin[y]}, {y, 0, 50}]
