I probably put a little bit too much effort into this. The only rational point on this curve is $(0,0)$ (as well as the points at infinity $(1 : 5 : 0)$ and $(1 : -5 : 0)$).

There's a slightly non-obvious change of variables that will turn your equation into $y^{2} = x^{5} + 2000^{2}$. Using this form of the equation, we can lift the points on the curve to one of a family of étale $5$-fold covers. In particular, if we write $x = a/b$ and $y = c/d$ with $\gcd(a,b) = \gcd(c,d) = 1$, then we get
$$ \frac{c^{2}}{d^{2}} = \frac{a^{5} + 2000^{2} b^{5}}{b^{5}}. $$
It's easy to see that the fraction on the right is reduced and so $c^{2} = a^{5} + 2000^{2} b^{5}$ and $d^{2} = b^{5}$. Thus, $b = n^{2}$ for some positive integer $n$. This gives
$$c^{2} - 2000^{2} n^{10} = (c - 2000 n^{5})(c + 2000 n^{5}) = a^{5}.$$
The two factors on the left hand side can only have the prime factors $2$ and $5$ in common, and their product is a $5$th power. Hence, $c + 2000n^{5} = 2^{\alpha} 5^{\beta} e^{5}$ for some integers $0 \leq \alpha, \beta \leq 4$. Letting $f = a/e$, we get
$$ 2^{2 \alpha} 5^{2 \beta} e^{5} = f^{5} + 2^{\alpha + 5} 5^{\beta + 3} n^{5}. $$
Each of these $25$ curves has genus $6$, and every rational point on the original curve lifts to one of these $25$ curves. The only ones that have local points are the three with $(\alpha,\beta) = (0,0), (0,2)$ and $(0,3)$. The curve with $(\alpha,\beta) = (0,0)$ is $a^{5} + b^{5} = 125c^{5}$, and the curves with $(\alpha,\beta) = (0,2)$ and $(0,3)$ are both isomorphic to $a^{5} + b^{5} = 625c^{5}$. The three rational points on the original curve all arise from the point $(1 : -1 : 0)$ on these curves. It suffices to prove there are no more points on these curves, in particular that it's not possible to write $125$ and $625$ as a sum of two rational $5$th powers.

I'll do this by using Frey curves. (There's a more elementary way to do the case of $a^{5} + b^{5} = 125c^{5}$, but the genus $2$ quotient of $a^{5} + b^{5} = 625c^{5}$ *also* has a Jacobian with $\mathbb{Q}$-rank $2$.)

Suppose that $a$, $b$ and $c$ are integers with $c \ne 0$ and $\gcd(a,b,c) = 1$ so that $a^{5} + b^{5} = 5^{\beta} c^{5}$ with $\beta \in \{3, 4 \}$. It's easy to see that $5 \nmid a$ and $5 \nmid b$. First assume that $c$ is odd. Swap $a$ and $b$ if necessary to force $a$ to be odd, and negate $a$ if necessary to make $a^{5} \equiv -1 \pmod{4}$. Then, $b$ is even. Let $E$ be the elliptic curve
$$E : y^{2} = x(x-a^{5})(x+b^{5}).$$
The discriminant of this curve is $16 \cdot 5^{2 \beta} a^{10} b^{10} c^{10}$, although there is a change of variables that will lower the discriminant and show that $E$ has multiplicative reduction at $2$. The $j$-invariant of $E$ is $\frac{(5^{2 \beta} c^{10} - a^{5} b^{5})^{3}}{5^{2 \beta} a^{10} b^{10} c^{10}}$.

Thinking about the mod $5$ Galois representation attached to $E$ will lead to a contradiction. This representation is irreducible. (All the $2$-torsion points on $E$ are rational, and if the mod $5$ representation was reducible, this would give a rational point on the modular curve parametrizing elliptic curves with full $2$-torsion and reducible mod $5$ Galois representation. This modular curve has genus $1$, and precisely $6$ rational points, all of which are cusps.)

Serre's conjecture (which is now a theorem) predicts that there is a modular form of weight $k$ and level $N$ whose mod $5$ Galois representation gives rise to that of $E$. The level $N$ is the product of the primes $\ell \ne 5$ for which the power $\ell$ dividing the discriminant is not a multiple of $5$. By the computation above, this is only $\ell = 2$, so $N = 2$. The weight of the modular form is $6 = 5+1$, because the power of $5$ dividing the discriminant is not a multiple of $5$. However, we get a contradiction because there are no weight $6$ cusp form of level $2$! (The lowest weight cusp form of level $2$ arises in weight $8$.) This contradiction proves that this elliptic curve doesn't exist, and so there are no solutions to $a^{5} + b^{5} = 5^{\beta} c^{5}$ with $c$ odd.

In the case that $c$ is even and $a$ and $b$ are odd, one can run the same argument starting with the elliptic curve $E : y^{2} = x(x-a^{5})(x-5^{\beta}c^{5})$ and the same things happen - the irreducible mod $5$ Galois representation arises in level $2$ and weight $6$, which is again a contradiction.