First given $x \in \mathbb{R}$:
$$\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2 + n}{n^2}=\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}$$
So if you prove $\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}$ is absolutly convergent you are ready (since the second sum converges and is constant and not depends on $x$).
$x^2$ is bounded for every finite interval on $I$ by $M_I$:
$$\left|(-1)^n\frac{x^2}{n^2}\right|\leq \frac{M_I}{n^2} $$
So $\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}$ is absolutly convergent.
Uniformity: pointwise $f_m(x)$ converges to:
$$f(x)=x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}$$
So:
$$|f(x)-f_m(x)|=$$
$$\left|x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-x^2 \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
$$\leq\left|x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}-x^2 \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
$$\leq|x^2|\left| \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}- \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
$$\leq M_I\left| \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}- \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
At that point you can complete the demostration (because the convergence is not given by the point $x$ you take)
To clarify your confusion:
Weierstrass M-test states one and only one thing:
Given a sequence of functions $f_k(x)$ defined on $E\subseteq\mathbb{R}$, the series $\sum f_k(x)$ converges uniformly if there exists a sequnece of reals $M_k$ such that $|f_k|\le M_k$ for each $k$ and $\sum M_k$ converges.
Note here that $M_k$ must not depend on $x$. This only means that if you found such sequence $M_k$ then the series uniformly converges. It says nothing about the series if you have found some $M_k$ whose series does not converge.
In particular, this test cannot be used to prove that some series does not converge uniformly.
Now consider
$$
f_k(x)=\frac{1}{kx+2}-\frac{1}{(k+1)x+2}
$$
Consider the partial sum
\begin{align*}
S_n(x)=\sum_{k=0}^{n} f_k(x)&=\left(\frac{1}{2}-\frac{1}{x+2}\right)+\cdots+\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right)\\
&=\frac{1}{2}-\frac{1}{(n+1)x+2}
\end{align*}
the series $\sum f_k(x)$ converges uniformly if and only if $S_n$ converges uniformly.
Now $S_n(0)=0$ for all $n$. So $S_n(0) \to 0$, and $S_n(x) \to \frac{1}{2}$ for $0<x\le 1$. So define
$$
S(x)=
\begin{cases}
0 & \text{if} \quad x=0 \\
\frac{1}{2} & \text{if} \quad 0<x\le 1
\end{cases}
$$
Then $S_n(x)\to S(x)$. Also, $S_n(0)-S(0)=0$ for any $n$. Now
$$
m_n:=\sup_{x\in [0,1]} |S_n-S|=\sup_{x\in (0,1]} \left\lvert -\frac{1}{(n+1)x+2} \right\rvert=\frac{1}{2} \not\to 0
$$
So $S_n$ does not converge uniformly.
Best Answer
You have to prove that your upper bound is valid. Use the fact that $e^{x} \geq x$ for $x >0$ (which follows from the Taylor expansion). Now $e^{n^{2}x^{2}} \geq e^{n^{2}a^{2}} \geq n^{2}a^{2}$ so $\frac 1 {e^{n^{2}x^{2}}} \leq \frac 1 {n^{2}a^{2}}$, now compare with the series $ \sum \frac 1 {n^{2}a^{2}}$ which is convergent.