My solution : Consider $p$ is a prime number such that $p \mid t+1$ and define $\alpha = v_p (t+1)$. Define $k=\alpha ! \phi(p^{\alpha+1}) $ Define $n= (t(t+1)+1)^k$ now $n^a+t \equiv t+1 (mod p^{\alpha +1})$ so $p^\alpha \mid \mid n^a+t$ so if $n^a+t= b^m $ then $m \leq \alpha$ now we know that $n^a =( (t(t+1)+1)^{\frac{ak}{m}})^m$ because $m \mid \alpha ! \mid k$ . now use this trivial lemma : $(x+1)^k-x^k \geq 2x+1 $ $(k \geq 2 , k,x \in \mathbb{N})$

Two other solutions are attached .The first one is like goldeneagle's solution and the second one has Arefs's solution as a part of itself. They are in Persian(Farsi).

MahyarSefidgaran wrote: Two other solutions are attached .The first one is like goldeneagle's solution and the second one has Arefs's solution as a part of itself. They are in Persian(Farsi). Second solution in English: First case: $t+1$ is not a perfect power. Define $n=t(t+1)^2+1$. Now consider $\exists k \in \mathbb{N} $ such that $n^k+t $ is a perfect power.Define $y=t(t+1)^2$ $(y+1)^k +t=y^k+ky^{k-1}+....+ky+t+1 = (t+1)(b(t+1)+1)$ and now Define$z= b(t+1)+1$ . Now $(z,t+1)=1$ so $t+1$ is perfect power. Contradiction!!! Second case: $t+1$ is a perfect power . Consider $t+1 = m^r $ and $m$ is not perfect power. Define $n_0= t(t+1)^2+1 , n=n_0^r$. Consider $n^k+t = c^d$ so by proof of first case $t+1= l^d$ so $d \mid r$ so $t=c^d - n^k= c^d - n_0 ^{kr} = c^d - n_0^{ksd} = (c-n_0)(c^{d-1}+....n_0^{ks(d-1)}) \geq n_0 > t$ Contradiction!!!!