Let me illustrate the approach I sketched in the comments to find a solution:

Let $\pi = 1+\sqrt{-5}$ be a uniformizer of $K=\mathbb{Q}_2(\sqrt{-5})$, and $v$ the valuation with $v(2)=1$, so $v(\pi) = \frac{1}{2}$. In order for $17x^4 + y^2 = -1$, the minimum valuation among the three terms has to occur twice, so there are three cases to study: $2v(x)=v(y)<0$, $v(x)=0\leq v(y)$ and $v(y)=0\leq v(x)$. In the latter two cases, we have to solve your equation in $\mathcal{O}_K$, and it suffices to do so modulo $\pi^9$ due to Hensel lifting (modulo $\pi^5$ in the last case).

In the first case, we may also turn this into an equation in $\mathcal{O}_K$ by multiplying with a suitable power of $\pi$. Let $n = -2v(x)$, then $\pi^n x, \pi^{2n}y \in \mathcal{O}_K^\times$, and multiplying the original equation with $\pi^4$ we are led to the equation $17(\pi^n x)^4 + (\pi^{2n}y)^2 = -\pi^{4n}$. Let us write this as $17u^4 + v^2 = -\pi^{4n}$. By Hensel lifting (in $v$), or equivalently the observation that $(1+\pi^5\mathcal{O}_K)\subseteq (\mathcal{O}_K^\times)^2$, it suffices to solve this equation modulo $\pi^5$. If $n\geq 2$, this means that we are looking for a solution of $17u^4 + v^2 = 0$ with units $u$ and $v$, which doesn't exist since $-17$ is not a square in $\mathcal{O}_K$. If $n=1$, we have that $\pi^4 = 4$ and $17 = 1$ modulo $\pi^5$, so we are asking for a solution to $x^4 + y^2 = -4$, and this does exist: $x=1$ and $y=\sqrt{-5}$.

(If we had not found a solution here, we would also have to analyze the other two cases from the introduction, these can of course be done in the same way.)

Let me illustrate the approach I sketched in the comments to find a solution:

Let $\pi = 1+\sqrt{-5}$ be a uniformizer of $K=\mathbb{Q}_2(\sqrt{-5})$, and $v$ the valuation with $v(2)=1$, so $v(\pi) = \frac{1}{2}$. In order for $17x^4 + y^2 = -1$, the minimum valuation among the three terms has to occur twice, so there are three cases to study: $2v(x)=v(y)<0$, $v(x)=0\leq v(y)$ and $v(y)=0\leq v(x)$. In the latter two cases, we have to solve your equation in $\mathcal{O}_K$, and it suffices to do so modulo $\pi^9$ due to Hensel lifting (modulo $\pi^5$ in the last case).

In the first case, we may also turn this into an equation in $\mathcal{O}_K$ by multiplying with a suitable power of $\pi$. Let $n = -2v(x)$, then $\pi^n x, \pi^{2n}y \in \mathcal{O}_K^\times$, and multiplying the original equation with $\pi^4$ we are led to the equation $17(\pi^n x)^4 + (\pi^{2n}y)^2 = -\pi^{4n}$. Let us write this as $17u^4 + v^2 = -\pi^{4n}$. By Hensel lifting (in $v$), or equivalently the observation that $(1+\pi^5\mathcal{O}_K)\subseteq (\mathcal{O}_K^\times)^2$, it suffices to solve this equation modulo $\pi^5$. If $n\geq 2$, this means that we are looking for a solution of $17u^4 + v^2 = 0$ with units $u$ and $v$, which doesn't exist since $-17$ is not a square in $\mathcal{O}_K$. If $n=1$, we have that $\pi^4 = 4$ and $17 = 1$ modulo $\pi^5$, so we are asking for a solution to $u^4 + v^2 = -4$, and this does exist: $u=1$ and $v=\sqrt{-5}$.

So the solution to the original equation is found in the vicinity of $(\pi^{-1}, \sqrt{-5}\cdot \pi^{-2})$ with Hensel lifting.

(If we had not found a solution here, we would also have to analyze the other two cases from the introduction, these can of course be done in the same way.)