Don Zagier's own explanation is here, page 4+5 (in German, my translation).

Consider the elliptic curve given by the equation $y^2=x(x+n)(x-n)$.
If $P=(x,y)$ is an arbitrary nontrivial solution (meaning $y\neq 0$)
of this equation, the point $P'=(x',y')$ constructed using the
Diophantine tangent method has the property, that not only the product
$x'(x'+n)(x'-n)$ ($=y'^2$), but also all three factors $x'$, $x'+n$,
$x'-n$ are squares, hence $n$ is congruent.

For the original solution
$P$ the numbers $x$ and $x\pm n$ need not be squares, but they are
strongly restricted up to quadratic factors. If for example $n$ is
prime and $\equiv 5$ (mod 8), then one can easily show that each of
these three numbers is of the form $\pm\square$, $\pm 2\cdot\square$,
$\pm n\cdot\square$ or $\pm 2n\cdot\square$ (with $\square$ a
rational square). This leads to the consideration of a finite number
of cases, that have to be examined one by one.

If for example
$x=-A^2$, $x+n=B^2$, $x-n=-C^2$ (and in our case of $n$ prime,
$n\equiv 5$ (mod 8) one can easily show that if there is any solution
if must be of this form), then we need to solve the set of equations
$C^2-B^2=2A^2$, $C^2-A^2=n$. The first equation can be immediately
solved using the Diophantine method: it must hold that
$A=2RS/M$, $B=(R^2-2S^2)/M$, $C=(R^2+2S^2)/M$ for suitable integers
$R,S,M$. In this way the problem is reduced to the solvability of $M^2
> n=R^4+4S^4$. For $n=5$ the solution is evidently $M=R=S=1$
($\Rightarrow x=-4$, $y=6$, $x'=6\frac{97}{144}$). For other prime
numbers $n$ one must occasionally repeat the descent one or more
times, when in the first step one applies the Diophantine method to
the quadratic equation $n=U^2+4V^2$ and tries to find a solution with
$UV=\square$ ($\Rightarrow U=R^2/M$, $V=S^2/M$). For $n=157$ this
method leads to a solution in a few steps.

This is quite remarkable,
as Fermat himself might have said, but the page is unfortunately too
small to record the solution: The three rational squares have in
numerator and denominator each almost 100 digits.

isincluded in Heegner's original application!) $\endgroup$ – Noam D. Elkies Nov 10 '15 at 19:05