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.

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.
