Proving Stewart's theorem without trig Stewart's theorem states that in the triangle shown below,
$$ b^2 m + c^2 n = a (d^2 + mn). $$

Is there any good way to prove this without using any trigonometry? Every proof I can find uses the Law of Cosines.

First of all, see the Wikipedia article Stewart's Theorem.  I personally prefer the symmetric formulation of the theorem, because it is easy to remember without any mnemonic trickery:

Let $A$, $B$, $C$ be points on a directed line $l$ in the Euclidean plane, and $P$ be a point anywhere in the plane. Then
  $$
PA^2\cdot BC + PB^2\cdot CA + PC^2\cdot AB + BC\cdot CA\cdot AB = 0~,
$$
  where the length of a line segment $XY$ on the line $l$ is positive if the segment has the same direction as the line, and is negative if it has the opposite direction.

Proof.  The proof is by straightforward calculation.
Choose a Cartesian coordinate system with the directed line $l$ as the first axis,
and let the points involved have the coordinates $A=(a,0)$, $B=(b,0)$, $C=(c,0)$, $P=(x,y)$. Stewart's identity is essentially just the polynomial identity
$$
(1) \quad (x-a)^2(c-b) + (x-b)^2(a-c) + (x-c)^2(b-a) + (c-b)(a-c)(b-a) = 0~,
$$
which is easily verified. (Go and really do the calculation: it's fun to observe everything cancel out.) Now add to the above identity the obvious identity
$$
y^2(c-b) + y^2(a-c) + y^2(b-a) = 0~,
$$
and you have the full-blown Stewart's theorem. Done.
Note that the theorem is true also in partly degenerate situations, where some of the points $A$, $B$, $C$ coincide, or the point $P$ is on the line $l$.
A geometric reasoning has to discuss these degenerate cases separately, while the computational proof smoothly subsumes them in the general case.
Remark. The identity $(1)$ is a generic polynomial identity, which means that it is an identity in the ring $\mathbb{Z}[a,b,c,x]$ of polynomials in (formal) variables $a$, $b$, $c$, $x$ with integer coefficients, and therefore instantiates to an identity when the variables are replaced by elements of an arbitrary commutative ring. This observation tells us that Stewart's theorem is valid in the 'Euclidean plane geometry' over any commutative ring, in particular over any field.
