Let $\zeta$ be a primitive 8'th root of unity. Assuming $x_1$ and $x_2$ have no common roots (which you certainly want to assume or else there's a trivial solution), the factors in
$$
x_1^4+x_2^4 = (x_1-\zeta x_2)(x_1-\zeta^3 x_2)(x_1-\zeta^5 x_2)(x_1-\zeta^7 x_2)
$$
are relatively prime in $\mathbb{C}[t]$, so the only way to get a square factor in $x_1^4+x_2^4$ is for at least one of the factors $x_1-\zeta^k x_2$ to have a square factor.

So now suppose (WLOG) $x_1-\zeta x_2 = y^2z$. If you require $x_1,x_2\in\mathbb{Q}[t]$, then we can take complex conjugates to get $x_1-\overline \zeta x_2 = \overline y^2\overline z$, which lets us solve
$$
x_1=\frac{\overline \zeta y^2 z-\zeta\overline y^2\overline z}{\overline\zeta-\zeta}
\quad\text{and}\quad
x_2=\frac{y^2z-\overline y^2\overline z}{\overline\zeta-\zeta}.
$$
Of course, if you choose $y$ and $z$ to have real coefficients, you get $x_1=1$ and $x_2=0$. But if you choose $y$ to satisfy $\overline y^2\ne y^2$, then you'll get an answer to your question, at least over $\mathbb C$.

ADDENDUM: Alex seems to have posted something similar while I was typing, but I'll post my answer, too.

2nd ADDENDUM: Take $y\in\mathbb{Q}(i)[t]$ and $z=1$, then I think you get a solution in $\mathbb{Q}(\sqrt2)[t]$.