Are there any rational solutions to this equation? I am not sure if this is an appropriate question, but I was asked this by a colleague today and do not know how to answer it.
1) Are there any rational solutions to the following equation:
$$x^3-8x^2+5x+1 = -7y^2(x-1)x$$
2) Is it possible that this is an elliptic curve in disguise? I have noticed that after projectivizing, there are two points at infinity. Perhaps this is okay under some change of variables? (I plead ignorance on this.)
 A: 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:A: [Complete revamp of answer. It is based on the one before, but is better!]
In Tim's hyperelliptic equation, make the change of variables $y$ to $y/(-7)^2$, and $x$ to $x/(-7)$, to get:
$$y^2=x(x+7)(x^3+56x^2+245x-343)$$
For every prime $p$ with $v_p(x)<0$, the valuation must in fact be even, thus appears to an even power in the factorisation of each of the factors on the right (as non-zero rational numbers).
We need only consider the squarefree parts of the factors. Assume $p$ divides the numerator of at least two of the factors, to an odd power. By a small gcd calculation (3 in fact), we see that $p$ must be $7$.
Thus, we must have that each factor is a square times a number in $\{ -7, -1, 1, 7 \}$. For each triple $(a,b,c)$ of numbers in the set, with $abc$ a square, there is a possible element of the 2-Selmer group defined by the curve $C_{a,b,c}$ (in $\mathbb{P}^4$):
$$x = au^2,$$ $$\\ x+7=bv^2,$$ $$x^3+56x^2+245x-343=cw^2$$
Some of these might not have points locally and will not define an element of the 2-Selmer. But we will not make any explicit local computations.
For each such triple, the curve $C_{a,b,c}$ has a morphism into each of the curves:
$$C_{a,b,c}^1:\\ y^2=c(x^3+56x^2+245x-343)$$
$$C_{a,b,c}^2:\\ y^2=acx(x^3+56x^2+245x-343)$$
$$C_{a,b,c}^3:\\ y^2=bc(x+7)(x^3+56x^2+245x-343)$$
A sage computation shows that for each such triple $(a,b,c)$, at least one of these three curves has no rational points (other than ones at infinity, or points that don't correspond to solutions of the original equation, i.e. $x=1$ or $x=7$).
Therefore, the original equation has no rational solutions.

Here is the sage code of the computation:
def cubic_to_ellipticcurve(f):
    a, l = f.coeffs()[-1], f.coeffs()
    return EllipticCurve([0,l[2],0,l[1]*a,l[0]*a^2])

def quartic_to_ellipticcurve(f):
    for fac in factor(f):
        if fac[0].degree() == 1:
            r = fac[0].roots()[0][0]
            v = f.variables()[0]
            f = f(v+r)
            f = sum([f.coeffs()[4-i]*v^i for i in [0..3]])
            return cubic_to_ellipticcurve(f)
    return None

R.<x> = QQ[]

possible_sels = []
for a in [-7,-1,1,7]:
    for b in [-7,-1,1,7]:
        c = a*b
        E1 = cubic_to_ellipticcurve(c*(x^3+56*x^2+245*x-343))
        if E1.rank() == 0 and E1.torsion_order() == 1:
            continue

        E2 = quartic_to_ellipticcurve(a*c*x*(x^3+56*x^2+245*x-343))
        if E2.rank() == 0 and E2.torsion_order() == 1:
            continue

        E3 = quartic_to_ellipticcurve(b*c*(x+7)*(x^3+56*x^2+245*x-343))
        if E3.rank() == 0 and E3.torsion_order() == 1:
            continue

        possible_sels += [(a,b,c)]

print possible_sels # prints []

A: As Dror says, according to Sage it's a genus 2 curve.  Homogenizing your equation gives a degree 4 projective plane curve $C$, with arithmetic genus thus $(d-1)(d-2)/2 = 3\cdot 2/2 = 3$.  The geometric genus is the arithmetic genus minus a contribution for each singular point.  There is exactly one singular point on this projective curve, and assuming it is simple we get genus 2.  Drawing a picture of this singular point, indeed it looks simple.  So the assertion that the projective genus is 2, which is output from Sage, is reasonable.  If indeed the genus is 2, then this implies that your equation has only finitely many rational solutions (by Faltings's theorem). 
There is no known algorithm to find all rational points on a curve of genus $\geq 2$ in general.  But one can always look.  So I wrote a little compiled Cython program to search for rational points $(x,y,z)$ on the homogenous equation, with $-B\leq x,y,z < B$ and $z>0$, and in 13 seconds found no points with $B=1000$.
I published the Sage worksheet where I did the above computations here.
