Why are these two upper bounds the same? I'm trying to upper bound the quantity $\int_t^{\infty} e^{-x^2} dx$ as a function of $t$. I came up with two ways of doing this:

We have that $(x-t)^2 = x^2 -2tx + t^2 \ge 0 \implies -x^2 \le -2tx + t^2$, so that $\int_t^{\infty} e^{-x^2} dx \le \int_t^{\infty} e^{-2tx}e^{t^2} dx= e^{t^2} \int_t^{\infty} e^{-2tx} dx = e^{t^2} \left(-\frac{1}{2t} e^{-2tx}\bigg\rvert_{x=t}^{x=\infty}\right) = \frac{1}{2t}e^{-t^2}$.

We have, on the domain $x \in [t, \infty)$, that $x \ge t \implies 1 \le \frac{x}{t}$, so that
$\int_t^{\infty}e^{-x^2} dx \le \int_t^{\infty} \frac{x}{t} e^{-x^2} dx = \frac{1}{t} \left(-\frac{1}{2}e^{-x^2} \bigg \rvert_{x=t}^{x=\infty}\right) = \frac{1}{2t} e^{-t^2}$.


Is it a coincidence that these two totally different methods give the same upper bounds? Or is there a connection between these two estimates that I'm not realizing?
The coincedence is that from the point of view of an integral your estimations are the same. Why?
Let's recall the substitution rule for integrals:
For $\varphi \in C^1([a,b])$ it holds
$$\int_{a}^{b} f(\varphi(x)) \cdot \varphi'(x)\,\mathrm{d}x = \int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x$$
If we now have a closer look we can see that both of your estimations are of the form of the LHS:
For $$f(x) = \frac{1}{2}\exp(-tx)$$ and $$\varphi_1(x) = 2x-t$$ we get $$\int_{t}^{\infty} f(\varphi_1(x)) \cdot \varphi_1'(x)\,\mathrm{d}x = \int_t^\infty \exp(-2tx+t^2) \,\mathrm{d}x$$ for your first estimator while taking
$$\varphi_2(x) = \frac{x^2}{t}$$ it gives us $$\int_{t}^{\infty} f(\varphi_2(x)) \cdot \varphi_2'(x)\,\mathrm{d}x = \int_t^\infty \frac{x}{t}\exp(-x^2) \,\mathrm{d}x$$ so your second estimator.
And because it holds $$\varphi_1(t) = \varphi_2(t) = t$$ as well as $$\varphi_1(\infty) = \varphi_2(\infty) = \infty$$ both functions give the same result by using the substitution rule namely $$\int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x = \int_t^\infty \frac{1}{2}\exp(-tx) \,\mathrm{d}x$$
Moreover: Any suitable $\varphi \in C^1([t,\infty))$ with $\varphi(t) = t$ and $\varphi(\infty) = \infty$ will lead to the same result.
So what you did in both cases was actually estimating $\exp(-x^2)$ by $f(\varphi(x)) \cdot \varphi'(x)$ for a suitable $\varphi$
You could also try and use the gaussian integral and taylor series to get that:
$$\int_t^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}+\sum_{n=0}^\infty\frac{(-1)^{n+1}t^{2n+1}}{(2n+1)n!}$$
or if you take the first few terms you get:
$$I\approx\frac{\sqrt{\pi}}{2}+\frac{t^3}{3}-t+O(-t^5)$$