Usually $k[x,y]/(p(x))$ and $k[y]$ are not isomorphic. For example, if $k = \mathbb{Q}$ or $\mathbb{R}$ and $p(x)$ is an irreducible polynomial of degree at least $2$, then these rings are different -- say, if $p(x) = x^2+1$, then here $k[x,y]/(p(x))$ is $k[y][i]$, where $i^2 = -1$, not just $k[y]$. There is a natural map involving $k[y]$ here, but it is the inclusion map $k[y] \to k[x,y]/(p(x))$ giving $k[x,y]/(p(x))$ as a finitely generated module over $k[y]$ (specifically, if $p(x)$ has degree $d$, then $1, x, x^2, ..., x^{d-1}$ is a module basis). In other words, if $L = k[x]/(p(x))$ is the field extension of $k$ by a root of $p(x)$, then $k[x,y]/(p(x))$ is $L[y]$, not $k[y]$.
With this change, your answer is correct: if $\mathbb{m}$ is a maximal ideal of $k[x, y]/(p(x)),$ then the ideal $(p(x), \mathbb{m})$ of $k[x, y]$ is maximal. The maximal ideals $\mathbb{m}$ of $k[x,y]/(p(x))$, however, aren't just the principal ideals generated by irreducible polynomials $g(y) \in k[y]$, because those polynomials may factor over the field $L = k[x]/(p(x))$. In the same example as before, $g(y) = y^2+1$ is irreducible in $k[y]$, but factors as $(y+x)(y-x) = y^2 - x^2 = y^2 + 1$ in $k[x,y]/(x^2+1)$. Instead, if $q(x,y)$ is an irreducible factor of $g(y)$ in $L[y] = (k[x]/(p(x)))[y] = k[x,y]/(p(x))$, then the ideal $(p(x), q(x,y))$ is maximal in $k[x,y]$.
Best Answer
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M \subset F[x] \tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) \tag 2$
for some
$m(x) \in F[x]; \tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $\mu$ of $m(x)$, satisfying as it does $\mu \ne 0$, is a unit; thus $\mu^{-1} m(x)$ is monic and
$(\mu^{-1} m(x)) = (m(x)); \tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), \; p(x), q(x) \in F[x], \; \deg p(x), \deg q(x) \ge 1; \tag 5$
consider the ideal
$(p(x)) \subsetneq F[x]; \tag 6$
it is clearly proper: since $\deg p(x) \ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$ other than $0$. Also,
$(m(x)) = (p(x)q(x)) \subsetneq (p(x)), \tag 7$
for
$p(x) \notin (p(x)q(x)) \tag 8$
lest for some
$r(x) \in F[x] \tag 9$
we have
$p(x) = r(x)p(x)q(x), \tag{10}$
or
$p(x)(r(x)q(x) - 1) = 0, \tag{11}$
whence
$r(x)q(x) = 1, \tag{12}$
which yields
$\deg r(x) + \deg q(x) = \deg 1 = 0, \tag{13}$
impossible in light of the assumption $\deg q(x) \ge 1$; thus we have shown that
$(m(x)) = (p(x)q(x)) \subsetneq (p(x)) \subsetneq F[x] \tag{14}$
in the event that $m(x)$ is reducible, which further shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.