It would be true, indeed, that if $\gamma$ and $\gamma^2+1$ satisfy the same minimal polynomial, then $K(\gamma)\cong K(\gamma^2+1)$ are isomorphic, hence have the same degree over $K$. Since $K(\gamma^2+1)\subseteq K(\gamma)$, the equality of degree would suffice to show equality of fields.
But that is not always the case. For example, if $K=\mathbb{Q}$ and $\gamma=\sqrt[15]{2}$, then notice that $\sqrt[15]{4}+1$ does not satisfy the same minimal polynomial as $\gamma$ (neither does $\sqrt[15]{4}$).
Instead, note that $K(\gamma)$ is an extension of $K(\gamma^2+1)$, since $\gamma^2+1\in K(\gamma)$. Now notice that $\gamma$ satisfies the polynomial $x^2 -(\gamma^2+1) +1 \in K(\gamma^2+1)[x]$, so the degree of $K(\gamma)$ over $K(\gamma^2+1)$ is at most $2$. Can it be equal to $2$?
More generally:
Proposition. Let $K$ be a field, and let $u$ be algebraic over $K$. If $[K(u):K]$ is odd, then $K(u)=K(u^2)$.
(I know two proofs: a slick one using Dedekind's Product Theorem, and a direct computational one.)
You can deduce what you want after noting that $K(\gamma^2+1) = K(\gamma^2)$.