Must an injective or surjective map in an infinite dimensional vector space be a bijection?

No, it does not. Consider the space $\ell_\infty$ of bounded sequences of real numbers. The map

$$S:\ell_\infty\to\ell_\infty:\langle x_0,x_1,x_2,\ldots\rangle\mapsto\langle 0,x_0,x_1,x_2,\ldots\rangle$$

that shifts each sequence one term to the right and adds a leading $0$ term is linear, injective, and clearly not surjective.

No it does not, take for example the linear operator $D$, the derivative map from

$$C^\infty(\Bbb R)\to C^\infty(\Bbb R)$$

Then this map is clearly surjective by the FTC, but it is not injective as the derivative of any constant is $0$, so the null-space is non-trivial.

Another counter-example, taken from algebra:

Let $K$ be a field, and consider the infinite dimensional $K$-vector space of polynomials in one indeterminate, $V=K[X]$. Multiplication by $X$ is a linear map, which is injective, but not surjective, since polynomials with a non-zero constant term are not attained.

It does not.

Take the space of the real sequences, and the transformation

$T(u) = v$ where $v_n = u_{n+1}$

Then $T$ is surjective, but it's not injective.

Abstract example: take a basis $(b_i)_{i\in I}$ of the space. Let be $J\subset I$ with $|J| = |I|$ and $\sigma:I\longrightarrow J$ a bijection. The linear function defined by $$S(b_i) = b_{\sigma(i)},$$ is injective but not surjective while $$T(b_i) = b_{\sigma^{-1}(i)},i\in J,$$ $$T(b_i) = 0,i\in I\setminus J,$$ is surjective but not injective.