Rewrite original equation as: $$(a-b)((a-b)^2(a+b)^2 - 2) = c^2 + 1 \qquad (1)$$We split the problem in $2$ cases. Case 1: $a-b$ is even integer. In this case $a+b$ is also even integer. This implies that $LHS$ of $(1)$ is divisible by $4$. As a result $c^2+1$ must be divisible by $4$, which forces $c^2 \equiv 3 \pmod 4$, which is clearly impossible. Case 2 $a -b$ is odd integer. In this case $a+b$ is also odd integer. Note that square of odd integer is always $1 \pmod 4$. This implies that $(a-b)^2(a+b)^2 -2 \equiv 1 -2 \equiv 3 \pmod 4$. If $(a+b)^2(a-b)^2 -2 \ne 1, -1$, then if it is $3 \pmod 4$ it has a prime factor which is $3 \pmod 4$. This implies that $c^2 + 1$ is divisible by some prime $p$ with $p \equiv 3 \pmod 4$. It is well known that $c^2+1$ is divisible only by $2$ or odd prime, which is $1 \pmod 4$. We conclude that $(a-b)^2(a+b)^2 -2= 1, - 1$, which yields to some case checking , which can be easily done