Fourth Degree Pell Equation I got stuck at the following of my research problem:

Prove that only solution to equation $4b^2-3a^4=1$ for odd positive integers $a$, $b$ is $(1,1)$.

I made factorization -->$3a^4 = (2b-1)(2b+1)$
from here since 2b-1 and 2b+1 are coprime,
$$2b-1 = 3x^4, 2b+1 = y^4$$
or
$$2b-1 = x^4, 2b+1 = 3y^4$$
obviously first one has no solution from modulo 3. Second one yields $x^4+2 = 3y^4$ which is more specific than original equation.
Another approach that might be useful is to rewrite the original equation as $(2b)^2 - 3(a^2)^2 = 1$ , we get a recursive equation for values of $a^2$ by pell and it is sufficient to prove that there is no odd perfect squares other than 1 in this recursive sequence (a similar method is used here https://arxiv.org/pdf/1705.03011.pdf )
I am seeking an elementary solution which doesn't include algebraic NT. Thanks.
 A: Let $a$ and $b$ be positive integers  such that
$$4b^2-3a^4=1.\tag{1}$$
Then indeed, as you note
$$3a^4=4b^2-1=(2b+1)(2b-1),$$
where the greatest common divisor of the two factors divides their sum, which is $2$. Of course it is clear from $(1)$ that $3a^4$ is odd, and hence the two factors are coprime. This means that either
$$2b+1=3x^4\qquad\text{ and }\qquad 2b-1=y^4,$$
for some coprime integers $x$ and $y$, or that
$$2b+1=x^4\qquad\text{ and }\qquad 2b-1=3y^4,$$
for some coprime integers $x$ and $y$. In the latter case, reducing mod $3$ shows that $b\equiv2\pmod{4}$ and so $x^4\equiv2\pmod{4}$, a contradiction. In the former case we find that
$$3x^4-y^4=2.$$
Then in the ring $\Bbb{Z}[\sqrt{-2}]$, which is a UFD
$$3x^4=y^4+2=(y^2+\sqrt{-2})(y^2-\sqrt{-2}),$$
where again the two factors on the right hand side are coprime, and so
$$y^2+\sqrt{-2}=(1+\sqrt{-2})(a+b\sqrt{-2})^4,$$
for some coprime integers $a$ and $b$. Expanding the right hand side and comparing the coefficients of $\sqrt{-2}$ shows that
$$a^4-8a^3b-12a^2b^2+16ab^3+4b^4=1.$$
This is a Thue equation, for which efficient deterministic algorithms to find all solutions exist. PARI/GP tells me that they are the following pairs $(a,b)$:
$$(-1,-1),\qquad (-1,0),\qquad (0,1),\qquad (1,1).$$
Correspondingly either $y=0$ or $y=1$, and so the only valid solution is $(x,y)=(1,1)$.
A: Comment: May be this idea helps:
$4b^2-3a^4=1 \Rightarrow 3(b-a^2)(b+a^2)=(1-b)(1+b)$
We consider following cases:
Case 1:
$$\begin{cases} 3b-3a^2=1-b\\b+a^2=1+b\rightarrow 
 a^2=1\rightarrow a=\pm 1\end {cases}$$
which gives $b= 1$ for $a=+1$
Case 2:
$$\begin{cases} 3b+3a^2=1+b\\b-a^2=1-b\end {cases}$$
which gives no integer solution.
Case 3:
$$\begin{cases}3b-3a^2=1-b\rightarrow 2b-3a^2=1\\b^2+a^2=1-b \end{cases}$$
this system also has no integer solution.
We may consider other cases.
