number theory - Finding the error in a proof regarding solutions of the equation $a^2=1+61b^2$

This is a Pell equation with $n = 61$ so it has solutions in integers. First we make the change of variables $a=1+x$ and simplify, the equation becomes $x\left(x+2\right)=61b^2$. Suppose $61$ divides $x$ so $x=61\bar{x}$ which leads to $$2+61\bar{x}=\frac{b^2}{\overline{x}}\tag{1}$$ which leads to two possibilities. If $\overline{x} , $\overline{x}$ divides $b$ and $b=\overline{x}m$. If $\overline{x}>b$ , $\overline{x}=bm$ where $b=m\overline{m}$ so $ \overline{x}=m^2\overline{m}$.

After substituting $b=\overline{x}m$ in $(1)$ we find $2+61\overline{x}=\overline{x}m^2$ which implies that $\overline{x}$ divides $2$. Trying the possible values for $\overline{x}$ leads to a contradiction.

After substituting $ \overline{x}=m^2\overline{m}$ and $b=m\overline{m}$ in $(1)$ we find $2+61m^2\overline{m}=\overline{m}$ which implies that $\overline{m}$ divides $2$. Trying the possible values for $\overline{m}$ leads to a contradiction.

Now we can suppose $61$ divides $x+2$. So $x+2=61\overline{x}$. Substituting into the original equation gives $$61\overline{x}-2=\frac{b^2}{\overline{x}}\tag{2}$$ Considering The two possibilities as in the previous case leads to a similar contradiction. This shows that the substitution is not possible, but we know this is wrong since there exists an integer solution to the original equation which we know can be expressed in the form $1+x$ for some nonzero integer $x$.

I should say that this method only generates the trivial solution $a=1$ and $b=0$. But why doesn't it generate other solutions? What was the (wrong?) restrictive assumption that forced this method to generate only the trivial solution?

I also think this implies that the trivial solution is the only solution which shows that something must be wrong with this argument.

Answer

There is a problem at the point where I say that if $\bar{x} and $\bar{x}$ divides $b^2$ then $\bar{x}$ divides $b$ since $\bar{x}$ may have a prime factor with power more than one whereas this prime number is present in $b$ but raised to a smaller power while keeping the fact that $\bar{x}$ is smaller than $b$.