Yes, and indeed there are infinitely many rational points: the birationally equivalent Diophantine equation given by J.Ramsden in his partial answer to his own question, $$X + Y = Z + T, \phantom{and} XYZT = c,$$ was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$), who in 1749 obtained the rational curve of solutions $$X = 6\frac{cr(2r^4+c)^2}{(r^4-4c)(r^8+10cr^4-2c^2)}, \phantom{and} Y = -2\frac{r^8+10cr^4-2c^2}{3r^3(r^4-4c)},$$ $$Z = -3\frac{r^5(r^4-4c)^2}{2(2r^4+c)(r^8+10cr^4-2c^2)}, \phantom{and} T = \frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)}.$$ Inverting J.Ramsden's birational transformation $$(X,Y,Z,T) = \left( \frac{x-1}{z}, \frac{y+1}{z}, \frac{x+1}{z}, \frac{y-1}{z} \right)$$ then yields $$x = \frac{(r^8+2c^2)(r^8-44cr^4-2c^2)}{(r^8+10cr^4-2c^2)^2}, \phantom{and} y = \frac{7r^4+8c}{9r^4}, \phantom{and} z = \frac{-4(r^4-4c)(2r^4+c)}{3r(r^8+10cr^4-2c^2)}.$$

Euler's solution is unpublished and somewhat mysterious; he gave the formulas in a letter to Goldbach but didn't explain how he found them, writing only that he obtained the solution "endlich nach vieler angewandter Mühe" [at last, after applying much effort]. The curve is singular, and finding such a curve requires going somewhat beyond the usual manipulation of elliptic fibrations on K3 surfaces, which works only if for these surfaces will not work unless $c$ is a square (the case in which R.Kloosterman already found a nonsingular rational curve of solutions). I've lectured on Euler's surface and solution a number of times over the past few years; here's the latest iteration. Thanks to Franz Lemmermeyer for bringing Euler's letter to my attention.

Added later: Those The same lecture notes show on page 43 a somewhat simpler curve that I found on Euler's surface, which also yields a somewhat simpler curve of solutions of $(x^2-1)(y^2-1)=cz^4$: $$x = \frac{u^{12}+48cu^8-48c^2u^4+128c^3}{u^4(u^4-12c)^2}, \phantom{and} y = \frac{5u^4+4c}{3u^4-4c}, \phantom{and} z = \frac{4(u^4+4c)}{u(u^4-12c)}.$$ For $c=2$, we get at $u=2$ the solution $(26, 11/5, 6)$ which is almost as simple as the solution $(5/3,17,4)$ that Mark Sapir noted.

Added later yet: Here's a transcription of Euler's 1749 letter. See the last page.

3 Added the somewhat simpler solution curve; also some local fixes

Yes, and indeed they there are infinitely many rational points: the birationally equivalent Diophantine equation given by J.Ramsden in his partial answer to his own question, $$X + Y = Z + T, \phantom{and} XYZT = c,$$ (given by J.Ramsden in his partial answer to his own question) was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$), who in 1749 obtained the rational curve of solutions $$(X,Y,Z,T) X =$$ $$\left( 6\frac{cr(2r^4+c)^2}{(r^4-4c)(r^8+10cr^4-2c^2)}, \phantom{and} Y = -2\frac{r^8+10cr^4-2c^2}{3r^3(r^4-4c)},$$ $$Z = -3\frac{r^5(r^4-4c)^2}{2(2r^4+c)(r^8+10cr^4-2c^2)}, \frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)} phantom{and} T = \right). frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)}.$$ Inverting J.Ramsden's birational transformation $$(X,Y,Z,T) = \left( \frac{x-1}{z}, \frac{y+1}{z}, \frac{x+1}{z}, \frac{y-1}{z} \right)$$ then yields $$x = \frac{(r^8+2c^2)(r^8-44cr^4-2c^2)}{(r^8+10cr^4-2c^2)^2}, \phantom{and} y = \frac{7r^4+8c}{9r^4}, \phantom{and} z = \frac{-4(r^4-4c)(2r^4+c)}{3r(r^8+10cr^4-2c^2)}.$$

Euler's solution is unpublished and somewhat mysterious; he mentioned it gave the formulas in a letter to Goldbach but didn't explain how he found itthem, writing only that he obtained it the solution "endlich nach vieler angewandter Mühe" [at last, after applying much effort]. The curve is singular, and finding such a curve requires going somewhat beyond the usual manipulation of elliptic fibrations on K3 surfaces, which (as R.Kloosterman also noted) works only if $c$ is a square (the case in which R.Kloosterman already found a nonsingular curve of solutions). I've lectured on this Euler's solution a number of times over the past few years; here's the latest iteration. Thanks to Franz Lemmermeyer for bringing Euler's letter to my attention.

Added later: Those lecture notes show on page 43 a somewhat simpler curve that I found on Euler's surface, which also yields a somewhat simpler curve of solutions of $(x^2-1)(y^2-1)=cz^4$: $$x = \frac{u^{12}+48cu^8-48c^2u^4+128c^3}{u^4(u^4-12c)^2}, \phantom{and} y = \frac{5u^4+4c}{3u^4-4c}, \phantom{and} z = \frac{4(u^4+4c)}{u(u^4-12c)}.$$ For $c=2$, we get at $u=2$ the solution $(26, 11/5, 6)$ which is almost as simple as the solution $(5/3,17,4)$ that Mark Sapir noted.

2 deleted 1 characters in body; added 3 characters in body

Yes, and indeed they are infinitely many rational points: the birationally equivalent Diophantine equation $$X + Y = Z + T, \phantom{and} XYZT = c$$ (given by J.Ramsden in his partial answer to his own question) was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$), who in 1749 obtained the rational curve of solutions $$(X,Y,Z,T) =$$ $$\left( 6\frac{cr(2r^4+c)^2}{(r^4-4c)(r^8+10cr^4-2c^2)}, -2\frac{r^8+10cr^4-2c^2}{3r^3(r^4-4c)}, -3\frac{r^5(r^4-4c)^2}{2(2r^4+c)(r^8+10cr^4-2c^2)}, \frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)} \right).$$ Inverting J.Ramsden's birational transformation $$(X,Y,Z,T) = \left( \frac{x-1}{z}, \frac{y+1}{z}, \frac{x+1}{z}, \frac{y-1}{z} \right)$$ then yields $$x = \frac{(r^8+2c^2)(r^8-44cr^4-2c^2)}{(r^8+10cr^4-2c^2)^2}, \phantom{and} y = \frac{7r^4+8c}{9r^4}, \phantom{and} z = \frac{-4(r^4-4c)(2r^4+c)}{3r(r^8+10cr^4-2c^2)}.$$

Euler's solution is unpublished and somewhat mysterious; he mentioned it in a letter to Goldbach but didn't explain how we he found it, writing only that he found obtained it "endlich nach vieler angewandter MueheMühe" [at last, after applying much effort]. The curve is singular, and finding such a curve requires going somewhat beyond the usual manipulation of elliptic fibrations on K3 surfaces, which (as R.Kloosterman also noted) works only if $c$ is a square. I've lectured on this solution a number of times over the past few years; here's the latest iteration. Thanks to Franz Lemmermeyer for bringing Euler's letter to my attention.

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