Assume the contrary. Let $(a, b)$ be a pair of natural numbers for which this does not hold, such that $a < b$ and this pair has a minimal sum of $a+b$.

Lemma 1. $a^2 + b^2 + 1$ is divisible by $2ab + 1$ if and only if $(2a^2 + 1)^2$ is divisible by $2ab + 1$.

Proof. Since $(a^2, 2ab+1)=1$, then $a^2+b^2+1$ is divisible by $2ab+1$ if and only if $a^4+b^2a^2+a^2$ is divisible by $2ab+1$ if and only if $4a^4+4b^2a^2+4a^2$ is divisible by $2ab+1$, since $2ab+1$ is odd. The rest follows from:
$$4a^4+4b^2a^2+4a^2 = (4a^4+4a^2+1)+(4b^2a^2-1) = (4a^4+4a^2+1)+(2ab-1)(2ab+1)$$
So the lemma is proved.

Thus for the fixed pair of natural numbers $(a, b)$ we have $a^2 + b^2 + 1$ is divisible by $2ab + 1$, therefore $(2a^2 + 1)^2$ is divisible by $2ab + 1$. So $(2a^2 + 1)^2 = (2ab + 1)(2ac + 1)$. It follows from the condition $a < b$ that $0 < c < a < b$, otherwise, the right-hand side is greater than the left-hand side. If $2ab + 1$ is not a perfect square then neither is $2ac + 1$. It follows from the lemma that if $(2a^2 + 1)^2$ is divisible by $2ac + 1$, then $a^2 + c^2 + 1$ is divisible by $2ac + 1$ as well. Thus we have found a pair of natural numbers that satisfies the problem statement and has a sum of components $c+a < a+b$, which yields the contradiction with the choice of $(a, b)$.