Let $a = 2+\sqrt{3}.$ By analogy to complex numbers let R$(a)$ be $r,$ the non-surd part of $r + s\sqrt{3}.$
I would like to show that a necessary but by no means sufficient condition that
$$(1)\hspace{10mm}R(a^{n+1}) \equiv 0~~~ \text{(mod 4n+3)} $$
is that $n = 6k+1$ with $k = 0, 1,2,...$
It would be enough to show by induction that $n = 6k,6k+2,6k+3,6k+4,6k+5$ are not congruent to $0$ mod $4n+3,$ so that any solutions obey the condition above. Beginning with $6k,$
$P(k=1)$: $~~R(a^{6\cdot1+1}) = 5042 \equiv 2~ (\text{mod 4(6k)+3})= 2~(\text{mod}~ 27).$
And
$P(k):$
$R(a^{6k+1})\not\equiv 0 ~~(\text{mod}~~ 4(6k)+3) = 24k + 3$
$R((a^6)^k\cdot a \not\equiv 0 ~(\text{mod}~ 24k+3)$
$?\implies$
$P(k+1):$
$R(a^{6(k+1)+1}\not\equiv 0(\text{mod}~24(k+1)+3)= 24k+27.$
$R((a^6)^k\cdot a^7)\not\equiv 0~(\text{mod}~24k+27).$
I am not really sure this can be done. Suggestions for doing the induction or better ideas appreciated.