<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] Here's a proof sketch. It's missing two major steps Propositions 1 and 2, but I'll try to fill them in and come back with an update.
First consider $x$ and $y$ even. Then we can only have $x\equiv 4\pmod{16}$, $y\equiv 8\pmod{16}$ and hence that $d=3$, for which @André Nicolas showed that $x=y=2$ is the only solution.
Now consider $y$ odd.
$d$ must be odd, so we can rewrite the equation as
$$
yz^2-x^2=4
$$
When $y=a^2+4$ then this has solutions
$$
\begin{array}{l}
z_0=z_1=1,z_{k+2}=(y-2)z_{k+1}-z_k \\
x_0=-a,x_1=a,x_{k+2}=(y-2)x_{k+1}-x_k
\end{array}
$$
Proposition 1: These enumerate all solutions, and there are no solutions in integers when $\sqrt{y-4}$ is not an integer.
Now to find a solution to the original problem we would need $z_i$ to be a power of $y$ for some $i$.
We can look at $z_k$ modulo $y$ and modulo $y^2$:
$$
\begin{align}
z_k & \equiv (-1)^{k+1}(2k-1) \pmod{y}\\
z_k & \equiv (-1)^{k}(2k-1)\left(\frac{k(k-1)}{6}y-1\right) \pmod{y^2}\\
\end{align}
$$
(These can be shown by induction.)
Thus $y\mid z_k$ iff $y\mid 2k-1$ and the first $z_k$ divisible by $y$ is when $k=(y+1)/2$.
Proposition 2: If $y\mid z_k$ then $z_{(y+1)/2} \mid z_k$.
Taking $z_{(y+1)/2}$ mod $y^2$:
$z_{(y+1)/2} \equiv -y\left(\frac{y^2-1}{24}y-1\right) \equiv -yw \pmod{y^2} $
where $\gcd(y,w)=1$. Hence $z_{(y+1)/2}$ is never divisible by $y^2$ and is either equal to $y$ or has a factor greater than 1 that does not divide $y$.
Since $z_k$ is increasing in $k$ and $z_3 = y^2-5y+5$ we find that $z_3=5$ when $y=5$ and $y<z_3<z_{(y+1)/2}$ when $y>5$. Thus $z_{(y+1)/2}$ is not a power of $y$ for $y>5$, and assuming Prop 2 nor can any other $z_k$ be, so a solution can only exist if $y=5$.
If $y=5$, $z_3=5,d=3,x=11$ is one solution.
The case $y=5$ is special in that $z_k=F_{2k-1}$ where $F_n$ is the $n$th Fibonacci number. We can use the divisibility properties of $F_n$ to show that if $25\mid z_k$ then $3001 \mid F_{25} \mid z_k$ so $z_k$ is not a power of 5. (This also shows Prop 2 is true when $y=5$ and hints how it might be proved in general.)
Thus the only solutions are $(x,y,d)=(\pm 2,2,3)$ or $(x,y,d)=(\pm 11,5,3)$.
Edit: Proposition 1 is incorrect, there may be solutions when $\sqrt{y-4}$ is not an integer, e.g. $39^2 = 61\cdot5^2-4$. This looks to be fatal to the argument.
</s>