Does an elementary solution exist to $x^2+1=y^3$? Prove that there are no positive integer solutions to $$x^2+1=y^3$$
This problem is easy if you apply Catalans conjecture and still doable talking about Gaussian integers and UFD's. However, can this problem be solved using pre-university mathematics?
I am talking about elementary number theoretical solutions. Do they exist?
 A: An alternative proof and a generalisation
As stated in the accepted answer, it is sufficient to prove that $3a^4+3a^2+1=b^2$ has  no positive integer solution. With a change of notation we shall prove the following generalisation by using the same method used for Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?
Theorem
No equation of either of the forms
$$ Ax^4-6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ ax^4+3x^2y^2+cy^4=z^2,ac=3\tag{2}$$
has a positive integer solution.
Proof
First note that, for either equation, we can suppose that $x,y,z$ are pairwise coprime  since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation can occur.
Also note that precisely one of $A$ and $C$ is divisible by $3$. Without loss of generality we can suppose that $3$ is a factor of $C$ and that $A\in \{\pm 1\}.$ Then $z^2\equiv A\pmod 3$ and so $A=1$. Similarly, we can suppose $a=1$.
An equation of form (1)
$x^4-6x^2y^2-3y^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2-3y^2-z}{2}\right )\left (\frac{x^2-3y^2+z}{2}\right)=3y^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $y$, a contradiction.
Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=3$ and $y=uv$, with $u$ and $v$ coprime. Then $$au^4+cv^4=x^2-3y^2=x^2-3u^2v^2.$$ Therefore $au^4+3u^2v^2+cv^4=x^2$, $ac=3$, an equation of form (2).
It is important to note that the mapping $(x,y,z)\rightarrow (u,v,w)$ is invertible. Only one solution set can map to $(u,v,w)$ by this process.
An equation of  form (2)
Let $u,v,x$ be a pairwise coprime solution of  $ u^4+3u^2v^2+3v^4=x^2$ and let $t$ be the greatest common divisor of $v$ and $2$. Then $(U,V,W)=(\frac{2u}{t},\frac{v}{t},\frac{4x}{t^2})$ is a pairwise coprime solution of $U^4+12U^2V^2+48V^4=W^2$.
This can be rewritten, using completing the square, as $$\left (\frac{U^2+6V^2-W}{2}\right )\left (\frac{U^2+6V^2+W}{2}\right)=-3V^4.$$
The bracketed factors, $L$ and $M$,  are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=-3$ and $V=XY$, with $X$ and $Y$ coprime. Then $$aX^4-6X^2Y^2+cY^4=U^2,ac=-3,$$ an equation of form (1). Again, this mapping of solutions is invertible.
Conclusion
We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) leads to another positive integer solution $(X,Y,Z)$, where $$Y=\frac{y}{tuX}.$$
Any monotonically strictly decreasing series of positive integers must terminate and so the above process must lead to solutions with $tuX=1$.
It is now straightforward to plug $u=X=1$ and thus $U=2$ back into the equations to obtain   the contradiction
$$1-6Y^2-3Y^4=4.$$
