Claim 1: there is no $0$. Let $a_{i-1}=0$ for some $i$, we have $a_{i+1}^{a_i}=a_i$. Now we notice that if $a_i>1$ then $a_i$ is not a perfect $a_i$th power, and it has no solutions. If $a_i=1$ then $a_{i+1}=1$, then $a_{i+2}^1=1+1$ so $a_{i+2}=2$. But then $a_{i+3}^2=2+1$ and we have contradiction. If $a_i=0$ then $1=0$, contradiction. If $a_i=-1$ then $a_{i+1}=-1$, but then $a_{i+2}^{-1}=-2$. contradiction. if $a_i<-1$ then $a_{i+1}^{a_i}=a_i$, which means $a_{i+1}$ is not integer, contradiction. Claim 2: there is no $1$ Let $a_i=1$ for some $i$, then $a_{i+1}=1+a_{i-1}$. Note that $a_{i+2}^{1+a_{i-1}}=a_{i-1}+2$ by plugging in. if $a_{i-1}\leq -2$ then $a_{i+2}$ is not integer, contradiction. if $a_{i-1}=-1$ then $a_{i+1}=0$, wrong by claim 1. We also have $a_{i-1}\neq 0$ by claim 1 if $a_{i-1}\geq 1$ then $a_{i-1}+2$ is not a perfect $a_{i-1}+1$th power, contradiction. Claim 3: there are no negative numbers Let $a_i=-k$ where $k$ is negative. Then $\frac{1}{a_{i+1}^k}=a_{i-1}-k$. As $a_{i-1}-k$ is an integer and so is its inverse, we have $a_{i+1}^k=\pm 1$. So $a_{i+1}=-1$ (by claim 2). So $a_{i+2}^{-1}=-k-1$. As $-k-1$ is integer, we have $a_{i+2}=-1$ and $k=0$, contradiction. Now that there are only numbers greater than $1$. We let $a_i=2$ for some $i$ We note that $a_{i+2}^{a_{i+1}}=a_{i+1}+2$ When $a_{i+1}>2$, $a_{i+1}+2$ is not a perfect $a_{i+1}$th power. So $a_{i+1}=2$. Induct up to see $a_k=2$ for all $k\geq i$. Now $2^2=2+a_{i-1}$ so $a_{i-1}=2$. Induct down to see $a_k=2$ for all $k<i$. So if a $2$ appeared in the sequence than all terms are $2$. Finally if all numbers greater than $2$ then $a_{i+2}=\sqrt[a_{i+1}]{a_{i+1}+a_i}<a_i$ (which can be easily seen by $a_i^{a_{i+1}}>a_{i+1}+a_i$) As we have $a_{i+2}>a_i$ we know that the sequence strictly decreases on even terms So there will be a negative term eventually. Contradiction. Therefore the only sequence satisfying the constraint is the one with all terms being $2$.