The answer is Yes.
Theorem. The following are equivalent for any Hausdorff space $X$.
$X$ is compact.
$X^\kappa$ is Lindelöf for any cardinal $\kappa$.
$X^{\omega_1}$ is Lindelöf.
Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if $X$ is compact, then $X^\kappa$ is compact and hence Lindelöf.
So suppose that we have a space $X$ that is not compact, but $X^{\omega_1}$ is Lindelöf. It follows that $X$ is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets $U_0 \subset U_1 \subset \dots U_n \dots$ with the union $\bigcup\lbrace U_n \; | \; n \in \omega \rbrace = X$.
For each $J \subset \omega_1$ of size $n$, let $U_J$ be the set $\lbrace s \in X^{\omega_1} \; | \; s(\alpha) \in U_n$ for each $\alpha \in J \rbrace$. As the size of $J$ increases, the set $U_J$ allows more freedom on the coordinates in $J$, but restricts more coordinates. If $J$ has size $n$, let us call $U_J$ an open $n$-box, since it restricts the sequences on $n$ coordinates. Let $F$ be the family of all such $U_J$ for all finite $J \subset \omega_1$
This $F$ is a cover of $X^{\omega_1}$. To see this, consider any point $s \in X^{\omega_1}$. For each $\alpha \in \omega_1$, there is some $n$ with $s(\alpha) \in U_n$. Since $\omega_1$ is uncountable, there must be some value of $n$ that is repeated unboundedly often, in particular, some $n$ occurs at least $n$ times. Let $J$ be the coordinates where this $n$ appears. Thus, $s$ is in $U_J$. So $F$ is a cover.
Since $X^{\omega_1}$ is Lindelöf, there must be a countable subcover $F_0$. Let $J^*$ be the union of all the finite $J$ that appear in the $U_J$ in this subcover. So $J^*$ is a countable subset of $\omega_1$. Note that $J^*$ cannot be finite, since then the sizes of the $J$ appearing in $F_0$ would be bounded and it could not cover $X^{\omega_1}$. We may rearrange indices and assume without loss of generality that $J^*=\omega$ is the first $\omega$ many coordinates. So $F_0$ is really a cover of $X^\omega$, by ignoring the other coordinates.
But this is impossible. Define a sequence $s \in X^{\omega_1}$ by choosing $s(n)$ to be outside $U_{n+1}$, and otherwise arbitrary. Note that $s$ is in $U_n$ in fewer than $n$ coordinates below $\omega$, and so $s$ is not in any $n$-box with $J \subset \omega$, since any such box has $n$ values in $U_n$. Thus, $s$ is not in any set in $F_0$, so it is not a cover. QED
In particular, to answer the question at the end, it suffices to take any uncountable $\kappa$.
