The short answer is no: no, there isn't a way to calculate $L_x$. Let me expand.
You may be aware of the uncertainty relation between position and momentum, which can be written like this
$$ \langle [ \hat{x}, \hat{p}_x ] \rangle \geq i \hbar $$
What this means is that it's impossible to know precisely both the position and the momentum of a particle at a given point in time. If your particle is in some state for which the position is known very precisely, then the momentum of the particle must be very uncertain. That is to say, the particle doesn't have a precise, definite momentum --- the best we can do is talk about its average momentum and uncertainties in its momentum.
It turns out that there's a very similar relation for angular momentum, which looks like this
$$ \langle [ \hat{L}_x, \hat{L}_y ] \rangle \geq i \hbar \langle \hat{L}_z \rangle $$
Here (as before) the angle brackets denote 'expectation value' and the square brackets denote the commutator. What this is saying, essentially, is that for any system which has a non-zero z-component of angular momentum (for example, our hydrogen atom in the state $m = 1$), the quantities $L_x$ and $L_y$ are intrinsically uncertain. The system simply doesn't have a definite x- or y-component of angular momentum.
Note that the quantum number $m$ means the z-component of angular momentum (in units of $\hbar$). So the 'formula' for getting from the quantum number $m$ to the z-component of angular momentum --- namely, $L_z = m\hbar$ --- isn't really a formula at all, it's basically just a definition.
As such, there's simply no way that we could write $L_x$ or $L_y$ in terms of $m$, in the same way that if you gave me a randomly shaped box and told me its height $h$, there's simply no way I would be able to write the width in terms of it. The height of the box doesn't tell me anything about the width.
However, because of what I've argued above, we can actually make a stronger statement than just 'you can't write $L_x$ in terms of $m$'. The stronger statement is that we couldn't even find a number that characterised the x-component of angular momentum, because for so long as we know the z-component, the x- and y- components are uncertain. Hope this helps!
That you can only ever know one of the three components of angular momentum is best seen not through the uncertainty principle, but on the states themselves.
Since $[L_i,L_j] = \epsilon_{ijk} L_k$, the three momentum operators are pairwise not simultaneously diagonalizable (since simultaneous diagonalizability implies that the operators commute), meaning there is no eigenstate of an $L_i$ that would also be an eigenstate of any of the others. Hence you can only ever know one one angular momentum component for certain, since "knowing for certain" means "having an eigenstate".
Well, actually, the non-commutativity only says that there's no eigenbasis, but there could be some shared eigenvectors. However, one can check for the angular momentum operators that no such state - except the one with zero total angular momentum - exists.
To see this from the uncertainty principle, you have to look at what, actually, the relations say. We have the three relations
$$ \sigma_x \sigma_y \ge \lvert \langle L_z \rangle \rvert $$
$$ \sigma_z \sigma_x \ge \lvert \langle L_y \rangle \rvert $$
$$ \sigma_y \sigma_z \ge \lvert \langle L_x \rangle \rvert $$
where we have discarded the annoying $\frac{\hbar}{2}$ and set $\sigma_i \equiv \sigma_{L_i}$. Now, if we know the values of $L_z$, then $\sigma_z = 0$, and $\lvert \langle L_x \rangle \rvert = \lvert \langle L_y \rangle \rvert = 0$ follows immediately.
This implies that $\sigma_x$ and $\sigma_y$ are non-zero if the actual angular momenta do not identically vanish.
Best Answer
If the electron is confined to the $x-y$-plane, it's $z$-position is fixed, i.e. certain, and hence the $z$-momentum infinitely uncertain by the uncertainty relation.
That paragraph is trying to say that, if the magnitude of $\vec L$ is larger than $L_z$, then $\vec L$ is not fixed, and if angular momentum is not fixed, i.e. conserved, then the motion does not lie in a plane (perpendicular to it). This is a classical argument that is completely beside the point of quantum mechanics. The paragraph is completely non-sensical from a quantum mechanics view-point since it seems to treat $\vec L$ as some vector in space when only one of its three components even take definite values at the same time. (That there are other sources taking this approach does not make it less non-sensical, before anyone asks)
For an eigenstate of the operator $L_z$, one may always ask what the expectation values of $L_x$ and $L_y$ on it are. It follows from the uncertainty principle that the expectation values of $L_x$ and $L_z$ on an eigenstate of $L_z$ must vanish, since we have the three relations $$ \sigma_x \sigma_y \ge \frac{\hbar}{2} \lvert \langle L_z \rangle \rvert $$ $$ \sigma_z \sigma_x \ge \frac{\hbar}{2}\lvert \langle L_y \rangle \rvert $$ $$ \sigma_y \sigma_z \ge \frac{\hbar}{2}\lvert \langle L_x \rangle \rvert $$ for $\sigma_i$ the standard deviation of $L_i$. Now, in an eigenstate of $L_z$, $\sigma_z = 0$, so the absolute values of the averages of $L_x$ and $L_y$ are bounded by $0$ (given that $\sigma_i$ is always finite), and hence must be zero themselves.