<s>[INST] Solve $ x^2+4=y^d$ in integers with $d\ge 3$ 
Find all triples of integers $(x,y,d)$ with $d\ge 3$ such that $x^2+4=y^d$.

I did some advance in the problem with Gaussian integers but still can't finish it. The problem is similar to Catalan's conjecture.
NOTE: You can suppose that $d$ is a prime.
Source: My head
 [/INST] What follows only takes care of $d=3$. We use the Gaussian integer approach that you mentioned. The method can be used with other $d$, but some details of the calculation depend on $d$. Thus one can only deal with one $d$ at a time.
We first examine the case $x$ odd.  Factor $x^2+4$ in the Gaussian integers as $(x+2i)(x-2i)$. Since $x$ is odd, $x+2i$ and $x-2i$ are relatively prime.  If their product is a perfect cube, each must be a (Gaussian) perfect cube.  Here we are using the Unique Factorization Theorem for $\mathbb{Z}[i]$.  Some care needs to be taken, because of units that may appear in front of the prime factorization.  But since all four units are cubes, they give no difficulty.
Let $x+2i=(a+bi)^3$.  Expand, and compare real and imaginary parts. We get $a^3-3ab^2=x$ and $3a^2b-b^3=2$.  Since $3a^2b-b^3=b(3a^2-b^2)$, there are only the possibilities $b=\pm 1$, $b=\pm 2$. The case $b=1$ gives $a=\pm 1$. The case $b=-1$ gives no solution. The case $b=2$ also gives no solution, while $b=-2$ gives $a=\pm 1$.  The case $b=1$ gives the solution $x=2$ which we were not looking for, and $b=-2$, $a=-1$  gives the solution $x=11$.  
Next we deal somewhat more briefly with the case $x$ even. Let $x=2s$ and $z=2t$. Our equation becomes $(s+i)(s-i)=2t^3$. Note that $s$ and $t$ must be odd. By considering factorizations again, we get that $s+i$ must have shape $(1\pm i)(a+bi)^3$, and an analysis similar to the one above works. We get (for $(1+i)(a+bi)^3$) that $(a-b)(a^2+4ab+b^2)=1$, which forces $b=0$, $a=1$ or $b=-1$, $a=0$.  Thus the only positive solution of $s^2+1=2t^3$ has $s=1$, and therefore the only positive even $x$ such that $x^2+4$ is a perfect cube is given by $x=2$.
Remark: Let $d>3$ be odd.  We could, for any specific $d$, use exactly the same strategy.  Think for example of $d=5$, and the case $x$ odd.  Then from the fact that $x+2i$ must be a perfect (Gaussian) fifth power, we get $x=a^5-10a^3b^2+5ab^4$ and $2=5a^4b-10a^2b^3-b^5$. The second equation forces $b=\pm 1$ or $b=\pm 2$. For these four possibilities we can quickly find any integer solution $a$ by using the "Factor Theorem." If  we get an integer solution $a$, that gives a solution of the original equation, and if we don't, we know the original equation has no solution.
For any fixed odd $d$, the kind of calculation described above can be carried out mechanically and quickly.  And it is clear from the analysis that for any fixed $d$, there can be at most finitely many solutions. But unfortunately the strategy can only take care of $d$'s one at a time.  
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