Testing whether a die is biased, reasoning about the approach Question :
A die is marked on one side, and the number of times that the mark appears is
recorded.
The mark appears once from 25 rolls, is the die biased?

working
$H_0 : $ The die is not biased
$H_1 : $ The die is biased
For this note that the die should follow $\sim Bin(n, p)$ as $\sim Bin(25,
\frac{1}{6})$.
Testing at a $95\%$ significance level we need $P(x \leq 1)$, where $x = $
number of rolls with the mark atop. This is found from $P(0) + P(1)$
using:
\begin{aligned}
    P(0) = {25 \choose 0}\left( \frac{1}{6} \right)^{0}\left( 1 - \frac{1}{6} \right)^{25} \\
    P(1) = {25 \choose 1}\left( \frac{1}{6} \right)^{1}\left( 1 - \frac{1}{6} \right)^{24} \\
  \end{aligned}
Which gives
\begin{aligned}
    P(0) \approx 0.0104 \\
    P(1) \approx 0.0524
  \end{aligned}
Therefore the probability is $P(0) + P(1) = 0.0628 $, as we're testing at a
$95\%$ level we have
$$
0.0628 > 0.05
$$
Meaning that the probability of the observed outcome isn't less than $5\%$, and
therefore we don't reject $H_0$.

edit
As stated the probability has been conducted for a one tailed test yet I've not
been specific about whether I'm using two or one. So that I would change the
hypothesis to read as:
$H_0$: The die is fair.
$H_1$: The die is biased such that there are less results of the marked side
than there would be for a non biased die.
 A: I think most of the reasoning is valid.
Just in the end, we observe that the mark appears once, so the probability to observe it is $P(1) \approx 0.0524$. At that point, to compute $P(0)+P(1)$ is not meaningful to answer the question.
The question "the die is biased" isn't very clear, but one can assume that we want to construct a 95% confidence interval for the hypothesis, and that we take it two-sided. 
This means that the underlying random variable $X$ follows the distribution $Bin(25,1/6)$ under $H0$, and we want to find $x_0$ and $x_1$ such that:
$P(X \leq x_0)=0.025$ and $P(X \geq x_1)=0.025$ (two-sided confidence interval), so that $P(x_0 < X < x_1)=0.95$.
Without expliciting the confidence interval, we can simply observe that $P(X=1)>0.05$, so it cannot be part of the 5% tail.
If we want to explicit the confidence interval, since the distribution $Bin(25,1/6)$ takes discrete values, I would choose $x_0 = 0$, $x_1=8$, and one can check that:
 $P(X \leq 0)=P(X=0) \approx 0.01$, $P(X \geq 8) \approx 0.04$.
The conclusion is the same: We cannot reject the hypothesis $H0$.
EDIT: I plot $P(X=x)$ here. With the two-sided interval hypothesis, I will not reject the hypothesis if $1 \leq x \leq 7$.

[I also replaced 'symmetric' by 'two-sided' in my answer, because here, the binomial distribution is not symmetric.]
EDIT 2: I've just shown your edit. With this hypothesis: "The die is biased such that there are less results of the marked side than there would be for a non biased die.", then yes, your reasoning is correct!
A: I'd grade you A, if you re-worded your hypo as "the die is biased against the marked side" because you're testing one sided hypo. The current wording is "the die is biased" allows for bias for the marked side, i.e. asks for a two sided hypo. For non symmetric distribution like binomial the two sided hypo is tricky. 
The average is $np=25/6$, so how would you construct the tails here? I'd stick with one sided and submit it with an appropriate wording of the hypo
