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Take an elliptic curve over $\mathbb{Q}$: $$ y^2=x^3+ax+b$$ For some $a$ and $b$. Let $(x_i,y_i) $ be the $i$th solution to this equation. Is it possible that for sufficient $a$ and $b$, $x_i$ could be the $i$th prime number? The $y$ values can be any rational number.

Obviously I am not asking for explicit $a$ and $b$, but is there any result or theorem that disallows this?

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    How are the solutions $(x_i,y_i)$ indexed? – 2017-01-18
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    @hardmath I was thinking from 1, although it doesn't really matter. Any way is fine. – 2017-01-18
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    @QuestionAble Doesn't the finiteness of integer points on elliptic curves obviously prevent this? – 2017-01-18
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    @ErickWong The y value here does not have to be integral. Correct me if I'm wrong, but doesn't the 'finitely many integral points' only apply when both $x$ and $y$ are integral? – 2017-01-18
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    @QuestionAble I'm assuming you still want $y$ to be rational? (if $y$ is completely unconstrained it is trivially true since everything is a solution). If $a$ and $b$ are integers, then the RHS is an integer, so $y$ is an integer. If $a$ and $b$ are rational, then you can rescale the equation so that the RHS is an integer. In any case, you should specify the constraint on $y$ in the OP, rather than using the vague description "can be anything". – 2017-01-18
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    @QuestionAble Thanks for the edit! i've expanded my comment into an answer. – 2017-01-21

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Yes, this is impossible. I'll only discuss the case where $a$ and $b$ are rational: I'm pretty sure the irrational case reduces to two simultaneous elliptic curves by $\mathbb Q$-linearity.

Clearly if $a$ and $b$ are both integers, then this follows immediately from Siegel's theorem that an elliptic curve has finitely many points, since $x^3 + ax + b$ is an integer and its square root (if rational) must be integral.

Now suppose $a$ and $b$ are rational, with least common denominator $d$ (so that $ad, bd\in \mathbb Z$). If $(x_0,y_0)$ is a rational point on $y^2 = x^3 + ax + b$ with $x_0\in\mathbb Z$, then $(d^2 x_0, d^3 y_0)$ is an integer point on the elliptic curve $y^2 = x^3 + d^4a x + d^6b$, which has integer coefficients. So again there are only finitely many cases of the latter.

But let's suppose we want to broaden the question a bit, and allow $x$ to encode the primes in some other way (such as $x_i = 1/p_i$). Then there is still an argument to be made against this possibility. The rational points on an elliptic curve $E$ have the structure of a finitely-generated abelian group, whose elements grow roughly exponentially in height with each generator. The number of rational points of height $\le H$ is thus pretty small, only $O(\log^k H)$ where $k$ and the implied constant depend on the rank of $E$ and the size of the torsion group.

Since the number of primes up to $H$ is much, much greater, this is enough to rule out many reasonable potential encodings of the primes into $x$. But I don't know enough to really say it is completely impossible (I suspect it's true that "$x_i$ has denominator $p_i$" is impossible but I don't know a proof that excludes it).