I think Dror's nice argument can be made more transparent as follows. Suppose that $y^2=
x(x+7)(x^3+56x^2+245x-343)$ with $x,y$ in $\bf Q$ and $x$ not $0$ or $-7$. One sees easily that at a prime $p$ other than $7$ neither $x$ nor $x+7$ can have odd ord, either negative or positive. It follows
that $x^3+56x^2+245x-343$ is either a square or $7$(square) in $\bf Q$.
As Dror says, the first case can't occur. Suppose we're in the second. Write:
$x=7a$, $x^3+56x^2+245x-343=343b^2$, so that $x(x+7)=343c^2$ for some $c$ in $\bf Q$. Then:
$a$ is not $0$ or $-1$, $a(a+1)=7c^2$, and $b^2=a^3+8a^2+5a-1$. The second equation shows that $a$ is in the $7$-adic integers and is $0$ or $-1\pmod7$. Furthermore $a$ is a square or $7$(square) in the
$7$-adic integers. So $a$ can only be $0 \pmod 7$. But then $b^2 = -1 \pmod 7$, which can't happen.
EDIT: This is quite wrong. But I think Dror made the same errors; see the comment I
attached to his answer.
FURTHER EDIT: I'll write out a less computer-dependent version of Dror's excellent answer.
The curve Y^2=X^3-8(X^2)+5X+1 has good reduction at each prime other than 2 or 7, and
cuspidal reduction at 7. So its conductor is 49(power of 2), and the same is true of the twists:
(1)... The transformation X-->X+3 takes A to 7(Y^2)=X^3+X^2-16X-29 which is isomorphic to
Y^2=X^3+7(X^2)-794X-9947. X-->X-2 takes this to Y^2=X^3+X^2-800X-8359. So A has the
minimal Weierstrass model (0,1,0,-800,-8359), and B has the minimal model (0,-1,0,-800,8359). Looking up Cremona's tables for curves of conductor 49(a power of 2) we find that
A is isomorphic to 392b, that B is isomorphic to 784d, and neither has any finite rational
(2a)... Consider the curve 7(Y^2)=X(X^3-8(X^2)+5X+1). The transformation X-->1/X,Y-->Y/X^2
takes this to 7(Y^2)=X^3+5(X^2)-8X+1. X-->X+1 takes this in turn to 7(Y^2)=X^3+8(X^2)
+5X-1. Then X-->-X gives B. So the only finite rational point on our curve is (0,0).
(2b)... Next consider -7(Y^2)=(X-1)(X^3-8(X^2)+5X+1). X-->X+1 takes this to -7(Y^2)= X(X^3-5(X^2)-8X-1), and X-->-X takes this in turn to -7(Y^2)=X^4+5(X^3)-8(X^2)+X. Applying X-->1/X, Y-->Y/X^2 we get B again. So (1,0) is the only finite rational point on our curve.
THE PROOF: The conclusions of 1,2a and 2b give Speiser's result: there are no rational
points (x,y) on Y^2=-7(X)(X-1)(X^3-8(X^2)+5X+1) other than (0,0) and (1,0). For let (x,y) be
such a point. Then at any odd prime other than 7 the ord of x cannot be an odd number, either positive or negative, and the same holds for the ord of x-1. So each of x, x-1 and x^3-8(x^2)+5x+1 has one of the forms: square, -(square), 7(square), or -7(square). The conclusion of 1 above tells us that x^3-8(x^2)+5x+1 can only be a square or -(a square).
Suppose that x^3-8(x^2)+5x+1 is a square. Then x(x-1) is -7(square). So x is integral at
p=7 and congruent to 0 or 1 mod 7 in the 7-adics. If x were congruent to 1, x^3-8(x^2)+5x+1
would not be a square. So x is congruent to 0, and x-1, being congruent to-1, can only be
-(square). So x=7(square), and the same holds for x(x^3-8(x^2)+5x+1) contradicting the
conclusion of 2a.
Suppose finally that x^3-8(x^2)+5x+1 is -(square). Then x(x-1)=7(square). So x is integral
at p=7 and congruent to 0 or 1 mod 7 in the 7-adics. If x were congruent to 0, x^3-8(x^2)+5x+1 would not be the negative of a square. So x is congruent to 1, and can only be a square. Then
x-1 is 7(square), and (x-1)(x^3-8(x^2)+5x+1) is -7(square). This contradicts the conclusion of 2b.