Complex solutions for Fermat-Catalan conjecture

The Fermat-Catalan conjecture is that $a^m + b^n = c^k$ has only a finite number of solutions when $a, b, c$ are positive coprime integers, and $m,n,k$ are positive integers satisfying $\frac{1}{m} + \frac{1}{n} +\frac{1}{k} <1$. There are currently only 10 solutions known (listed at the end of this post). My question concerns the case where $a, b, c$ are positive coprime Gaussian integers. I've found two solutions. Is there a clever method for finding more? I've used brute force techniques.

• $(8+5i)^2+(5+3i)^3=(1+2i)^7$
• $(20+9i)^2+(1+8i)^3=(1+i)^{15}$
• $(1+2i)^7+(49+306i)^2=(27+37i)^3$ (Zander)
• $(44+83i)^2+(31+39i)^3=(5+2i)^7$ (Zander)
• $(238+72i)^3+(7+6i)^8=(7347−1240i)^2$ (Oleg567)

Here are the known solutions over integers.

• $1^m+2^3=3^2$
• $2^5+7^2=3^4$
• $13^2+7^3=2^9$
• $2^7+17^3=71^2$
• $3^5+11^4=122^2$
• $33^8+1549034^2=15613^3$
• $1414^3+2213459^2=65^7$
• $9262^3+15312283^2=113^7$
• $17^7+76271^3=21063928^2$
• $43^8+96222^3=30042907^2$

I found the next complex solution! :) $$(238+72i)^3+(7+6i)^8=(7347−1240i)^2$$

(There is no new method here. Just small contribution to the problem.)

Here is a small portion of what I have found:

$i^{4 m}+2^3=3^2$ where integer $m \ge 2$, i.e. the smallest is $i^8+2^3=3^2$

$(1+i)^2=i^{4 m+1}+i^{4 n+1}$ where integer $m \ge 1$, $n \ge 1$, i.e. the smallest is $(1+i)^2=i^5+i^5$

$(78-78 i)^2=(23 i)^3+i^{4 m+3}$ where integer $m \ge 1$, i.e. the smallest is $(78-78 i)^2=(23 i)^3+i^7$

$(11+11 i)^2+i^{4 m+1}=(3 i)^5$ where integer $m \ge 1$, i.e. the smallest is $(11+11 i)^2+i^5=(3 i)^5$

$(34-34 i)^2=(5 i)^3+(3 i)^7$

$(36+19 i)^2+(9-8 i)^3=(1+i)^{13}$