Let the given circle be $x^2 + y^2 - 4y + 3 = 0$. Rewrite it in standard form:

$x^2 + (y^2 - 4y) + 3 = 0$
$x^2 + (y - 2)^2 - 4 + 3 = 0$
$x^2 + (y - 2)^2 - 1 = 0$
$x^2 + (y - 2)^2 = 1$

So, the circle has centre $(0, 2)$ and radius $1$.

Let the centre of the required circle be $(h, k)$ and its radius be $r$.

Since the circle is tangent to the $x$-axis, its distance from the $x$-axis is $r$, so $k = r$.

Since the circle is externally tangent to the given circle, the distance between their centres is equal to the sum of their radii:

$\sqrt{(h - 0)^2 + (k - 2)^2} = r + 1$

But $k = r$, so:

$\sqrt{h^2 + (r - 2)^2} = r + 1$

Square both sides:

$h^2 + (r - 2)^2 = (r + 1)^2$
$h^2 + r^2 - 4r + 4 = r^2 + 2r + 1$

Subtract $r^2$ from both sides:

$h^2 - 4r + 4 = 2r + 1$
$h^2 - 4r + 4 - 2r - 1 = 0$
$h^2 - 6r + 3 = 0$

Recall $k = r$, so:

$h^2 - 6k + 3 = 0$

Therefore, the locus of the centres $(h, k)$ is:

$h^2 - 6k + 3 = 0$