Can $\operatorname{Re}(a+bi)^{n}$ be overlapped with $a,b\in\mathbb{Z}$ fixed?

Is there any integer solution for $$\operatorname{Re}((a+bi)^{m})=\operatorname{Re}((a+bi)^{n})$$ except $(m,n)=(0,1),(1,3)$, where $0\leq m<n,\ |a|\neq |b|,\ a\neq 0,\ b\neq 0$?

In other words,

Can $\operatorname{Re}(a+bi)^{n}$ be overlapped with $a+bi\in\mathbb{Z}[i]$ fixed except for some trivial cases?

This is a generalization of my earlier question, Is there any integer solution for $\operatorname{Re}(a+bi)^n=\pm1$, where $n\geq 2$, except $(a,b)=(\pm1,0),(0,\pm1)$?. The answer to this question is no. So, we have no solution for $m=0$.

I checked for every $0<|a|,|b|\leq 10000,\ |a|\neq|b|,\ 0\leq m<n\leq 1000$, then only found these:
$(a,b,m,n)=(\pm 2,\pm 1,1,3), (\pm 7,\pm 4,1,3), (\pm 26,\pm 15,1,3), (\pm 97,\pm 56,1,3), (\pm 362,\pm 209,1,3), (\pm 1351,\pm 780,1,3), (\pm 5042,\pm 2911,1,3)$

They are the integer solutions for $\operatorname{Re}(a+bi)^{1}=\operatorname{Re}(a+bi)^{3} \iff a^2-3b^2=1$.
I couldn't find any solutions for $(m,n)\neq (1,3)$.

PS
Just for your information, I also checked for $\operatorname{Im}(a+bi)^n$, then I found these solutions:
$(a,b,m,n)=(-2,\pm4,2,3),(8,\pm24,4,5),(9,\pm15,2,3),(-32,\pm56,2,3),(121,\pm209,2,3),(-450,\pm780,2,3),(1681,\pm2911,2,3)$

Except for $(a,b,m,n)=(8,\pm24,4,5)$, they are the integer solutions for $\operatorname{Im}(a+bi)^{2}=\operatorname{Im}(a+bi)^{3} \iff 3a^2b-b^3=2ab \iff 3a^2-2a=b^2$.

I would appreciate any help. Thank you for your cooperation.

This answer is to prove that there are no solutions for some particular values of $(m, n)$.

$m=1$, $n=4$:

In that case, the equation becomes $a = a^4 - 6a^2b^2 + b^4$, that is, $a(1+4ab^2) = (a^2-b^2)^2 $. It is easy to rule out $a = \pm b$. So suppose $a \neq \pm b$. Because $a$ and $1+4ab^2$ are coprime, $a$ and $1+4ab^2$ are $\pm$ squares. Suppose $a > 0$. Then $4ab^2$ and $1+4ab^2$ are squares. Thus $4ab^2=0$, a contradiction. Suppose $a < 0$. Then $-4ab^2$ and $-4ab^2-1$ are squares, so that $-4ab^2=1$, a contradiction.

$m=2$, $n = 3$:

In this case, $a^2-b^2 = a^3-3ab^2$. Rewrite it as $b^2(3a-1) = a^2(a-1) $. Because $a$ and $3a-1$ are coprime, $a^2 \mid b^2$, so that $a \mid b$. Then $3a-1 \mid a-1$. But $|3a-1| > |a-1|$ when $a \neq 0$, a contradiction.

$m = 2$, $n = 4$:

In this case, $a^2-b^2 = a^4 - 6a^2b^2 + b^4$. That is, $(2(a^2-b^2)-1)^2 = 16 a^2b^2+1$. Necessarily, $16a^2b^2 = 0$.

Case: $n$ odd and $m$ even

Writing $z=a+bi=re^{i\theta}$, the condition $\Re(z^n-z^m)=0$ gives $\cos m\theta=r^{n-m}\cos n\theta$.

When $r=\sqrt{a^2+b^2}$ is an integer, this forces $\cos\theta=p/q$ with $(p,q)=1$ so the rational root test can be used on the equation $$r^{n-m}T_n(\cos\theta)-T_m(\cos\theta)=0\tag1$$ where $T_\bullet$ denotes the Chebyshev polynomial of the first kind.

If $n$ is odd and $m$ is even, the roots must be of the form $\cos\theta=\pm1/q$ with $q\mid r$ as $a=r\cos\theta$ is an integer. Thus we have $a=\pm r/q$ and $b=\pm a\sqrt{q^2-1}\in\Bbb Z$. However, there is no integer $q>1$ such that $q^2-1$ is a square so no solutions can exist.

When $r$ is a non-integer, squaring $(1)$ and using the product identity for the Chebyshev polynomial yields $$r^{2(n-m)}T_{2n}(\cos\theta)-T_{2m}(\cos\theta)+r^{2(n-m)}-1=0\tag2$$ after rearrangement of terms. Notice that $r^2$ is necessarily an integer so $\cos^2\theta$ must be rational; the same equation can be used for the the rational root test as all powers of $\cos\theta$ are even.

If $n$ is odd and $m$ is even, the roots must be of the form $\cos^2\theta=p/q$ where $p=1,2$ and $q\mid r^2$. Simultaneously from the definition, we have $\cos^2\theta=a^2/(a^2+b^2)$ which automatically excludes $p=2$ since $(p,q)=1$. Thus $q=1+b^2/a^2\in\Bbb Z$ so $a\mid b$.