Statement: Suppose $K/k$ is a finite extension of finite fields. Show that both the norm and trace are injective.
Proof: For a finite extension of fields $K/k$, we associate to each element $\alpha$ of $K$ the $k$-linear transformation $m_\alpha: K\to K$ where $m_\alpha$ is multiplication by $\alpha$:
$m_\alpha(x)=\alpha x$ for $x\in K$. Each $m_\alpha$ is a $k$-linear function from $K$ to $K$. We define $$ Tr_{K/k}(\alpha) = Tr(m_\alpha)=Tr(\alpha x)=\alpha Tr(x). $$ Assume $\alpha, \beta \in K$ where $\alpha\neq \beta$. Then $$ Tr_{K/k}(\alpha)=Tr_{K/k}(\beta) \iff \alpha Tr(x) = \beta Tr(x) \iff \alpha =\beta $$ Hence the trace is injective.
We define $$ N_{K/k}(\alpha)=N(m_\alpha)=\det(\alpha x) =\alpha^n \det(x) $$ where $n=[K:k]$. Similarly, we find that the norm is also injective.
Now this seems too easy. Did I do something wrong?
The statement isn't true. The norm and trace take values in the ground field, so for example the image of $N_{\mathbb{F}_{49}/\mathbb{F}_7}$ must send a set of 49 elements to a set of 7, so it cannot be injective.
As for why your argument is wrong, you are thinking of $\alpha$ as a scalar when it is not, so you can't pull it out of the trace or determinant like you would an element of the ground field.