Let $M$ be any countably infinite set.
Let $X_{M}$:=$\{A\in M:|A|<\aleph_{0}\}$ (The set of all finite subsets of $M$) and
let $Y_{M}$:=$\{B\in M:B=M\backslash A, A\in X_{M}\}$ (The set of all countably infite subsets of $M$, whose complement is finite) and
let $Z_{M}$:=$\{C\in M:|C|=|M\backslash C|=\aleph_{0}\}$. (The set of all countably infite subsets of $M$, whose complement is countably infinite).
If $f$ is a function from $X_{M}$ to $Y_{M}$, such that $f(T):=M\backslash T$, then $f$ is bijective.
($\because$ If $f(U)=f(W)\implies M\backslash U=M\backslash W\implies U=W \implies f$ is one-one.
If $S\in Y_{M}$, then by definiton of $Y_{M}$, $M\backslash S \in X_{M}$, also $f(M\backslash S)=M\backslash (M\backslash S)=S \implies f$ is onto.)
Thus, $|X_{M}|=|Y_{M}|=\aleph_{0}$ and also we have $X_{M} \cup Y_{M} \cup Z_{M}=P(M)$ and $X_{M} \cap Y_{M} \cap Z_{M}=\varnothing \implies |Z_{M}|>\aleph_{0}$
Based, on this can we say that for any countably infinite set $M$, its power set $P(M)$is uncountable due to the existence of $Z_{M}$?
Based on your argument, no.
You're concluding that $Z_M$ is uncountable because the other two parts of the power set are countable. But unless you already know that $\mathcal P(M)$ is uncountable, why would that be true? If we divide $\Bbb Q$, the rational numbers, into the positive integers, the negative integers, and all the rest of them rational numbers, the first two parts are easily countable, can we conclude that the third part is uncountable? No.
If, however, your argument would have been a diagonalisation argument of some sort, or a direct argument that $Z_M$ is uncountable, then you can say that $\mathcal P(M)$ is uncountable, because you'd shown that it has an uncountable subset.