Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$.

Certainly, $x$ must be odd. Note that $(x^2+1)^2 = x^4+2x^2+1 = 2y^2+2x^2$ and $(x^2-1)^2 = x^4-2x^2+1 = 2y^2-2x^2.$ Multiply these two equations together to get

$$(x^4-1)^2 = 4(y^4-x^4).$$

Since $x^4\equiv 1 \pmod{4}$, we can write

$$\left(\frac{x^4-1}{2}\right)^2 = y^4-x^4.$$

This implies a solution to the better-known Diophantine equation

$$X^4-Y^4 =Z^2.$$

This equation can be proved to have only trivial solutions by infinite descent, so we must have $x^4-1 = 0$. Hence $x=\pm 1$ which forces $y=\pm 1$.

For the descent (switching to lower case):

We have $z^2 + (y^2)^2 = (x^2)^2.$ By the usual construction of solutions to the Pythagorean equation, either $y^2=2mn$ or $y^2=m^2-n^2.$ If $y^2 = 2mn,$ then $m = u^2,$ and $n = 2v^2.$ Then $z^2 = m^2 - n^2 = u^4 - v^4,$ and $u^4 = m^2 < m^2 + n^2 < x^2 < x^4,$ so we have a smaller positive solution. If $y^2 = m^2 - n^2,$ $z = 2mn,$ and $x^2 = m^2 + n^2,$ then $x^2y^2 = m^4 - n^4$ which is a smaller solution, because $m^2 < x^2.$

With $z=x^2$ we can consider $z^2-2y^2=-1$, which is a well studied Pell-like equation: $a^2-D\,b^2=-1$, where $D$ is a positive, nonsquare natural number. Solving it involves finding the continued fraction representation for $\sqrt D=\sqrt 2$, which is quite simple:

\begin{align}\sqrt 2 &= [1,2,2,2,\dots]. \end{align}

For these $D$, the representation is always eventually periodic and of the form

$$\sqrt D = [a_0,\overline{a_1,a_2,\dots,a_r,2a_0}].$$

When $r$ is odd, there is no solution, but when $r$ is even $($as in our case, with $r=0)$, the smallest (or fundamental) solution is given by $(a,b)=(p_r,q_r)$, where $p_n/q_n$ is the $n$-th convergent $[a_0,a_1,\dots,a_n]$ of $\sqrt D$.

Our case is all too simple, and $p_0=q_0=1$.

Once we have the fundamental solution, all other solutions can be obtained by taking $n$-th powers, where $n$ is odd. Indeed, from $p_r^2-Dq_r^2=-1$ it follows that $\left(p_r^2-Dq_r^2\right)^n=-1$. Settin $a^2-Db^2=\left(p_r^2-Dq_r^2\right)^n$, we can do some factoring and manipulations to arrive at

$$a=\frac{\left(p_r+\sqrt Dq_r\right)^n+\left(p_r-\sqrt Dq_r\right)^n}2$$ $$b=\frac{\left(p_r+\sqrt Dq_r\right)^n-\left(p_r-\sqrt Dq_r\right)^n}{2\sqrt D}$$

In our case, with $n=2k+1$ our solutions look like

\begin{align} z &=\frac{\left(1+\sqrt 2\right)^{2k+1}+\left(1-\sqrt 2\right)^{2k+1}}2\tag{$*$} \end{align}

These particular number have been studied before, and are called Newman-Shanks-Williams number, or NSW numbers; Our question hinges on: for which values of $k$ is $z$ a square number? Well, a simple application of Newton's binomial theorem shows that

\begin{align} z &=\sum_{\substack{0\leq i \leq 2k+1\\i\,\text{ even}}}\binom{2k+1}i\left(\sqrt{2}\right)^i\\ &=\sum_{0\leq j \leq k}\binom{2k+1}{2j}2^j \end{align}

The first few $z$ are $1,7,41,239, \dots$ and can be found on the OEIS. I am not quite managing to show that $z$ cannot be a square for $k>0$, so I might turn back on this later.

We consider $x, y$ positive (if $x$ is solution then $-x$ is solution, same with $y$). From $x^4+1=2y^2$ we deduce that $2y^2=t^2+1$, where $t=x^2$ must be odd. Let $t=2k+1$, so $y^2=k^2+(k+1)^2$. We can now use the fact that $y=(m+n)^2$ and $k=(m-n)^2$ and $(k+1)=2mn$ or $k=2mn$ and $(k+1)=(m-n)^2$, where $m, n\in \mathbb Z$.

Case 1. $k=(m-n)^2$ and $(k+1)=2mn$.

Then $k+2(k+1)=(m+n)^2=y$, so $y=3k+2$ so $$9k^2+12k+4=2k^2+2k+1$$ and we find k.

Case 2. $k=2mn$ and $(k+1)=(m-n)^2$

Then $2k+k+1=(m+n)^2=y$, so $y=3k+1$ so $$9k^2+6k+1=2k^2+2k+1$$ and we find k.

EDIT: It is NOT right, sorry for the confusion