Is the representation of any prime of the form $6n+1$ as $a^2+3b^2$ essentially unique? Two well-known results in number theory are:

Fermat's $4n+1$ theorem: Every prime of the form $4n+1$ can be represented as $a^2+b^2 (a,b \in\mathbb{N})$.


Euler's $6n+1$ theorem: Every prime of the form $6n+1$ can be represented as $a^2+3b^2 (a,b \in\mathbb{N})$.

Looking at the Mathworld entries on these theorems here and here, I notice that representation of primes of the form $4n+1$ is stated to be unique (up to order), but that there is no mention of uniqueness in respect of representation of primes of the form $6n+1$.  Uniqueness does however seem to hold at least for small primes of this form.
Question: Is the representation of any prime of the form $6n+1$ as $a^2+3b^2$ essentially unique?
If this is the case then a reference to a proof would be appreciated.

If you require $a, b \in \mathbb{N}$ then the representation is literally unique, and we can argue as Daniel Fischer does in the comments: the ring of Eisenstein integers $\mathbb{Z}[\omega]$, where $\omega = \frac{-1 + \sqrt{-3}}{2}$ is a primitive third root of unity, is a unique factorization domain, and the primes congruent to $1 \bmod 6$ admit a further factorization
$$p = (x - y \omega)(x - y \omega^2) = x^2 + xy + y^2$$
where $x - y \omega$ is a prime in $\mathbb{Z}[\omega]$ and $x - y \omega^2$ is its conjugate. (This can be proven by inspecting the quotient $\mathbb{Z}[\omega]/p \cong \mathbb{F}_p[\omega]/(\omega^2 + \omega + 1)$.) We can write $x - y \omega = x + \frac{y}{2} - \frac{y}{2} \sqrt{-3}$  which gives
$$p = \left( x + \frac{y}{2} \right)^2 + 3\left( \frac{y}{2} \right)^2$$
and this representation consists of integers as long as $y$ is even. Now, $x - y \omega$ in the factorization above is unique up to multiplication by units, and the units of $\mathbb{Z}[\omega]$ are $\pm 1, \pm \omega, \pm \omega^2$. Multiplying by $\pm 1$ doesn't affect the values of $|x|$ and $|y|$ so now we inspect only the results of multiplying by $\omega$ and $\omega^2$. This gives
$$(x - y \omega) \omega = x \omega - y \omega^2 = x \omega + y (\omega + 1) = y + (x + y) \omega$$
$$(x - y \omega) \omega^2 = x \omega^2 - y = - x (\omega + 1) - y = (- x - y) - x \omega.$$
This means that if $p = a^2 + 3b^2 = (a + \sqrt{-3} b)(a - \sqrt{-3} b)$ then $b$ can only take on one of the six values $\pm \frac{y}{2}, \pm \frac{x+y}{2}, \pm \frac{x}{2}$, and some casework gives that depending on the parities of $x$ and $y$, exactly one of $x, y, x + y$ is even (the case where $x$ and $y$ are both even can't occur because then $p$ would be divisible by $4$). So $b$ is (the absolute value of) this unique even value divided by $2$, which uniquely determines $a$.
